{"id":19689,"date":"2022-02-21T23:23:46","date_gmt":"2022-02-21T23:23:46","guid":{"rendered":"https:\/\/onlinemathtutor.co\/?page_id=19689"},"modified":"2022-05-19T20:51:02","modified_gmt":"2022-05-19T20:51:02","slug":"integration-practice-problems-ibp","status":"publish","type":"page","link":"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/integration-practice-problems-ibp\/","title":{"rendered":"Integration Practice Problems &#8211; [IBP]"},"content":{"rendered":"\t\t<div data-elementor-type=\"wp-page\" data-elementor-id=\"19689\" class=\"elementor elementor-19689\" data-elementor-post-type=\"page\">\n\t\t\t\t\t\t<section class=\"elementor-section elementor-top-section elementor-element elementor-element-513b2f27 elementor-section-boxed elementor-section-height-default elementor-section-height-default\" data-id=\"513b2f27\" data-element_type=\"section\">\n\t\t\t\t\t\t<div class=\"elementor-container elementor-column-gap-default\">\n\t\t\t\t\t<div class=\"elementor-column elementor-col-100 elementor-top-column elementor-element elementor-element-77ef05d8\" data-id=\"77ef05d8\" data-element_type=\"column\">\n\t\t\t<div class=\"elementor-widget-wrap elementor-element-populated\">\n\t\t\t\t\t\t<div class=\"elementor-element elementor-element-62931345 elementor-widget elementor-widget-heading\" data-id=\"62931345\" data-element_type=\"widget\" data-widget_type=\"heading.default\">\n\t\t\t\t<div class=\"elementor-widget-container\">\n\t\t\t\t\t<h1 class=\"elementor-heading-title elementor-size-default\">Practice Problems: Integation By Parts<\/h1>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"elementor-element elementor-element-592c0d33 elementor-nav-menu__align-center elementor-nav-menu--stretch elementor-nav-menu__text-align-center animated-fast elementor-nav-menu--dropdown-tablet elementor-nav-menu--toggle elementor-nav-menu--burger elementor-widget elementor-widget-nav-menu\" data-id=\"592c0d33\" data-element_type=\"widget\" data-settings=\"{&quot;submenu_icon&quot;:{&quot;value&quot;:&quot;&lt;i class=\\&quot;fas fa-pencil-alt\\&quot;&gt;&lt;\\\/i&gt;&quot;,&quot;library&quot;:&quot;fa-solid&quot;},&quot;full_width&quot;:&quot;stretch&quot;,&quot;_animation&quot;:&quot;none&quot;,&quot;layout&quot;:&quot;horizontal&quot;,&quot;toggle&quot;:&quot;burger&quot;}\" data-widget_type=\"nav-menu.default\">\n\t\t\t\t<div class=\"elementor-widget-container\">\n\t\t\t\t\t\t\t\t<nav aria-label=\"Menu\" class=\"elementor-nav-menu--main elementor-nav-menu__container elementor-nav-menu--layout-horizontal e--pointer-double-line e--animation-fast\">\n\t\t\t\t<ul id=\"menu-1-592c0d33\" class=\"elementor-nav-menu\"><li class=\"menu-item menu-item-type-custom menu-item-object-custom menu-item-19677\"><a href=\"https:\/\/onlinemathtutor.co\/index.php\/problem-set-integration\/\" class=\"elementor-item\"><img decoding=\"async\" src=\"https:\/\/onlinemathtutor.co\/wp-content\/uploads\/2021\/06\/Home-Icon-Blue.png\"  width=\"20\"><\/a><\/li>\n<li class=\"menu-item menu-item-type-custom menu-item-object-custom menu-item-has-children menu-item-19577\"><a href=\"#\" class=\"elementor-item elementor-item-anchor\">Integration Techniques<\/a>\n<ul class=\"sub-menu elementor-nav-menu--dropdown\">\n\t<li class=\"menu-item menu-item-type-custom menu-item-object-custom menu-item-has-children menu-item-19777\"><a href=\"#\" class=\"elementor-sub-item elementor-item-anchor\">U-Substitution<\/a>\n\t<ul class=\"sub-menu elementor-nav-menu--dropdown\">\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-19576\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/integration-practice-problems-usub\/\" class=\"elementor-sub-item\">Problem Set<\/a><\/li>\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-19776\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/virtual-lessons-integration-usub\/\" class=\"elementor-sub-item\">Virtual Lessons<\/a><\/li>\n\t<\/ul>\n<\/li>\n\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-has-children menu-item-19694\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/integration-practice-problems-ibp\/\" class=\"elementor-sub-item\">Integration By Parts<\/a>\n\t<ul class=\"sub-menu elementor-nav-menu--dropdown\">\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-19898\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/integration-practice-problems-ibp\/\" class=\"elementor-sub-item\">Problem Set<\/a><\/li>\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-19956\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/virtual-lessons-integration-ibp\/\" class=\"elementor-sub-item\">Virtual Lessons<\/a><\/li>\n\t<\/ul>\n<\/li>\n\t<li class=\"menu-item menu-item-type-custom menu-item-object-custom menu-item-has-children menu-item-19885\"><a href=\"#\" class=\"elementor-sub-item elementor-item-anchor\">Trigonometric Integrals<\/a>\n\t<ul class=\"sub-menu elementor-nav-menu--dropdown\">\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-19925\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/integration-practice-problems-trigonometric-integrals\/\" class=\"elementor-sub-item\">Problem Set<\/a><\/li>\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-19954\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/virtual-lessons-integration-trigonometric-integrals\/\" class=\"elementor-sub-item\">Virtual Lessons<\/a><\/li>\n\t<\/ul>\n<\/li>\n\t<li class=\"menu-item menu-item-type-custom menu-item-object-custom menu-item-has-children menu-item-19886\"><a href=\"#\" class=\"elementor-sub-item elementor-item-anchor\">Trigonometric Substitution<\/a>\n\t<ul class=\"sub-menu elementor-nav-menu--dropdown\">\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-19926\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/integration-practice-problems-trigonometric-substitution\/\" class=\"elementor-sub-item\">Problem Set<\/a><\/li>\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-19953\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/virtual-lessons-integration-trigonometric-substitution\/\" class=\"elementor-sub-item\">Virtual Lessons<\/a><\/li>\n\t<\/ul>\n<\/li>\n\t<li class=\"menu-item menu-item-type-custom menu-item-object-custom menu-item-has-children menu-item-19884\"><a href=\"#\" class=\"elementor-sub-item elementor-item-anchor\">Partial Fractions<\/a>\n\t<ul class=\"sub-menu elementor-nav-menu--dropdown\">\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-19924\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/integration-practice-problems-partial-fractions\/\" class=\"elementor-sub-item\">Problem Set<\/a><\/li>\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-19955\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/virtual-lessons-integration-partialfractions\/\" class=\"elementor-sub-item\">Virtual Lessons<\/a><\/li>\n\t<\/ul>\n<\/li>\n\t<li class=\"menu-item menu-item-type-custom menu-item-object-custom menu-item-has-children menu-item-19893\"><a href=\"#\" class=\"elementor-sub-item elementor-item-anchor\">Improper Integrals<\/a>\n\t<ul class=\"sub-menu elementor-nav-menu--dropdown\">\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-19927\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/integration-practice-problems-improper-integrals\/\" class=\"elementor-sub-item\">Problem Set<\/a><\/li>\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-19966\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/virtual-lessons-integration-improper-integrals\/\" class=\"elementor-sub-item\">Virtual Lessons<\/a><\/li>\n\t<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n<li class=\"menu-item menu-item-type-custom menu-item-object-custom menu-item-has-children menu-item-19578\"><a href=\"#\" class=\"elementor-item elementor-item-anchor\">Integration Applications<\/a>\n<ul class=\"sub-menu elementor-nav-menu--dropdown\">\n\t<li class=\"menu-item menu-item-type-custom menu-item-object-custom menu-item-has-children menu-item-19888\"><a href=\"#\" class=\"elementor-sub-item elementor-item-anchor\">Area<\/a>\n\t<ul class=\"sub-menu elementor-nav-menu--dropdown\">\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-19928\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/integration-practice-problems-application-area\/\" class=\"elementor-sub-item\">Problem Set<\/a><\/li>\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-20136\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/virtual-lessons-integration-area\/\" class=\"elementor-sub-item\">Virtual Lessons<\/a><\/li>\n\t<\/ul>\n<\/li>\n\t<li class=\"menu-item menu-item-type-custom menu-item-object-custom menu-item-has-children menu-item-19889\"><a href=\"#\" class=\"elementor-sub-item elementor-item-anchor\">Volumes [Nonrotational]<\/a>\n\t<ul class=\"sub-menu elementor-nav-menu--dropdown\">\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-19929\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/integration-practice-problems-application-volume\/\" class=\"elementor-sub-item\">Problem Set<\/a><\/li>\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-20582\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/virtual-lessons-integration-trigonometric-substitution\/\" class=\"elementor-sub-item\">Virtual Lessons<\/a><\/li>\n\t<\/ul>\n<\/li>\n\t<li class=\"menu-item menu-item-type-custom menu-item-object-custom menu-item-has-children menu-item-19890\"><a href=\"#\" class=\"elementor-sub-item elementor-item-anchor\">Volume [Rotational]<\/a>\n\t<ul class=\"sub-menu elementor-nav-menu--dropdown\">\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-19931\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/integration-practice-problems-application-rotational-volumes\/\" class=\"elementor-sub-item\">Problem Set<\/a><\/li>\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-20160\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/virtual-lessons-integration-rotational-volume\/\" class=\"elementor-sub-item\">Virtual Lessons<\/a><\/li>\n\t<\/ul>\n<\/li>\n\t<li class=\"menu-item menu-item-type-custom menu-item-object-custom menu-item-has-children menu-item-19891\"><a href=\"#\" class=\"elementor-sub-item elementor-item-anchor\">Surface Area<\/a>\n\t<ul class=\"sub-menu elementor-nav-menu--dropdown\">\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-19930\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/integration-practice-problems-application-surface-area\/\" class=\"elementor-sub-item\">Problem Set<\/a><\/li>\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-20165\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/virtual-lessons-integration-surface-area\/\" class=\"elementor-sub-item\">Virtual Lessons<\/a><\/li>\n\t<\/ul>\n<\/li>\n\t<li class=\"menu-item menu-item-type-custom menu-item-object-custom menu-item-has-children menu-item-19892\"><a href=\"#\" class=\"elementor-sub-item elementor-item-anchor\">Arc Length<\/a>\n\t<ul class=\"sub-menu elementor-nav-menu--dropdown\">\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-20177\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/integration-practice-problems-arc-length\/\" class=\"elementor-sub-item\">Problem Set<\/a><\/li>\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-20176\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/virtual-lessons-integration-arc-length\/\" class=\"elementor-sub-item\">Virtual Lessons<\/a><\/li>\n\t<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n<li class=\"menu-item menu-item-type-custom menu-item-object-custom menu-item-has-children menu-item-19579\"><a href=\"#\" class=\"elementor-item elementor-item-anchor\">Numerical Integration<\/a>\n<ul class=\"sub-menu elementor-nav-menu--dropdown\">\n\t<li class=\"menu-item menu-item-type-custom menu-item-object-custom menu-item-has-children menu-item-19894\"><a href=\"#\" class=\"elementor-sub-item elementor-item-anchor\">Reimann Sums<\/a>\n\t<ul class=\"sub-menu elementor-nav-menu--dropdown\">\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-20110\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/integration-practice-problems-riemann-sums\/\" class=\"elementor-sub-item\">Problem Set<\/a><\/li>\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-20128\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/virtual-lessons-integration-riemann-sum\/\" class=\"elementor-sub-item\">Virtual Lessons<\/a><\/li>\n\t<\/ul>\n<\/li>\n\t<li class=\"menu-item menu-item-type-custom menu-item-object-custom menu-item-has-children menu-item-19895\"><a href=\"#\" class=\"elementor-sub-item elementor-item-anchor\">Trapezoidal Rule<\/a>\n\t<ul class=\"sub-menu elementor-nav-menu--dropdown\">\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-20109\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/integration-practice-problems-trapezoidal\/\" class=\"elementor-sub-item\">Problem Set<\/a><\/li>\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-20138\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/virtual-lessons-integration-trapezoidal\/\" class=\"elementor-sub-item\">Virtual Lessons<\/a><\/li>\n\t<\/ul>\n<\/li>\n\t<li class=\"menu-item menu-item-type-custom menu-item-object-custom menu-item-has-children menu-item-19896\"><a href=\"#\" class=\"elementor-sub-item elementor-item-anchor\">Simpson&#8217;s Rule<\/a>\n\t<ul class=\"sub-menu elementor-nav-menu--dropdown\">\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-20108\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/integration-practice-problems-simpsons\/\" class=\"elementor-sub-item\">Problem Set<\/a><\/li>\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-20137\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/virtual-lessons-integration-simpsons\/\" class=\"elementor-sub-item\">Virtual Lessons<\/a><\/li>\n\t<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n<\/ul>\t\t\t<\/nav>\n\t\t\t\t\t<div class=\"elementor-menu-toggle\" role=\"button\" tabindex=\"0\" aria-label=\"Menu Toggle\" aria-expanded=\"false\">\n\t\t\t<i aria-hidden=\"true\" role=\"presentation\" class=\"elementor-menu-toggle__icon--open eicon-menu-bar\"><\/i><i aria-hidden=\"true\" role=\"presentation\" class=\"elementor-menu-toggle__icon--close eicon-close\"><\/i>\t\t<\/div>\n\t\t\t\t\t<nav class=\"elementor-nav-menu--dropdown elementor-nav-menu__container\" aria-hidden=\"true\">\n\t\t\t\t<ul id=\"menu-2-592c0d33\" class=\"elementor-nav-menu\"><li class=\"menu-item menu-item-type-custom menu-item-object-custom menu-item-19677\"><a href=\"https:\/\/onlinemathtutor.co\/index.php\/problem-set-integration\/\" class=\"elementor-item\" tabindex=\"-1\"><img decoding=\"async\" src=\"https:\/\/onlinemathtutor.co\/wp-content\/uploads\/2021\/06\/Home-Icon-Blue.png\"  width=\"20\"><\/a><\/li>\n<li class=\"menu-item menu-item-type-custom menu-item-object-custom menu-item-has-children menu-item-19577\"><a href=\"#\" class=\"elementor-item elementor-item-anchor\" tabindex=\"-1\">Integration Techniques<\/a>\n<ul class=\"sub-menu elementor-nav-menu--dropdown\">\n\t<li class=\"menu-item menu-item-type-custom menu-item-object-custom menu-item-has-children menu-item-19777\"><a href=\"#\" class=\"elementor-sub-item elementor-item-anchor\" tabindex=\"-1\">U-Substitution<\/a>\n\t<ul class=\"sub-menu elementor-nav-menu--dropdown\">\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-19576\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/integration-practice-problems-usub\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Problem Set<\/a><\/li>\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-19776\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/virtual-lessons-integration-usub\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Virtual Lessons<\/a><\/li>\n\t<\/ul>\n<\/li>\n\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-has-children menu-item-19694\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/integration-practice-problems-ibp\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Integration By Parts<\/a>\n\t<ul class=\"sub-menu elementor-nav-menu--dropdown\">\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-19898\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/integration-practice-problems-ibp\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Problem Set<\/a><\/li>\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-19956\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/virtual-lessons-integration-ibp\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Virtual Lessons<\/a><\/li>\n\t<\/ul>\n<\/li>\n\t<li class=\"menu-item menu-item-type-custom menu-item-object-custom menu-item-has-children menu-item-19885\"><a href=\"#\" class=\"elementor-sub-item elementor-item-anchor\" tabindex=\"-1\">Trigonometric Integrals<\/a>\n\t<ul class=\"sub-menu elementor-nav-menu--dropdown\">\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-19925\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/integration-practice-problems-trigonometric-integrals\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Problem Set<\/a><\/li>\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-19954\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/virtual-lessons-integration-trigonometric-integrals\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Virtual Lessons<\/a><\/li>\n\t<\/ul>\n<\/li>\n\t<li class=\"menu-item menu-item-type-custom menu-item-object-custom menu-item-has-children menu-item-19886\"><a href=\"#\" class=\"elementor-sub-item elementor-item-anchor\" tabindex=\"-1\">Trigonometric Substitution<\/a>\n\t<ul class=\"sub-menu elementor-nav-menu--dropdown\">\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-19926\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/integration-practice-problems-trigonometric-substitution\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Problem Set<\/a><\/li>\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-19953\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/virtual-lessons-integration-trigonometric-substitution\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Virtual Lessons<\/a><\/li>\n\t<\/ul>\n<\/li>\n\t<li class=\"menu-item menu-item-type-custom menu-item-object-custom menu-item-has-children menu-item-19884\"><a href=\"#\" class=\"elementor-sub-item elementor-item-anchor\" tabindex=\"-1\">Partial Fractions<\/a>\n\t<ul class=\"sub-menu elementor-nav-menu--dropdown\">\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-19924\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/integration-practice-problems-partial-fractions\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Problem Set<\/a><\/li>\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-19955\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/virtual-lessons-integration-partialfractions\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Virtual Lessons<\/a><\/li>\n\t<\/ul>\n<\/li>\n\t<li class=\"menu-item menu-item-type-custom menu-item-object-custom menu-item-has-children menu-item-19893\"><a href=\"#\" class=\"elementor-sub-item elementor-item-anchor\" tabindex=\"-1\">Improper Integrals<\/a>\n\t<ul class=\"sub-menu elementor-nav-menu--dropdown\">\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-19927\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/integration-practice-problems-improper-integrals\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Problem Set<\/a><\/li>\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-19966\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/virtual-lessons-integration-improper-integrals\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Virtual Lessons<\/a><\/li>\n\t<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n<li class=\"menu-item menu-item-type-custom menu-item-object-custom menu-item-has-children menu-item-19578\"><a href=\"#\" class=\"elementor-item elementor-item-anchor\" tabindex=\"-1\">Integration Applications<\/a>\n<ul class=\"sub-menu elementor-nav-menu--dropdown\">\n\t<li class=\"menu-item menu-item-type-custom menu-item-object-custom menu-item-has-children menu-item-19888\"><a href=\"#\" class=\"elementor-sub-item elementor-item-anchor\" tabindex=\"-1\">Area<\/a>\n\t<ul class=\"sub-menu elementor-nav-menu--dropdown\">\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-19928\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/integration-practice-problems-application-area\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Problem Set<\/a><\/li>\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-20136\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/virtual-lessons-integration-area\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Virtual Lessons<\/a><\/li>\n\t<\/ul>\n<\/li>\n\t<li class=\"menu-item menu-item-type-custom menu-item-object-custom menu-item-has-children menu-item-19889\"><a href=\"#\" class=\"elementor-sub-item elementor-item-anchor\" tabindex=\"-1\">Volumes [Nonrotational]<\/a>\n\t<ul class=\"sub-menu elementor-nav-menu--dropdown\">\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-19929\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/integration-practice-problems-application-volume\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Problem Set<\/a><\/li>\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-20582\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/virtual-lessons-integration-trigonometric-substitution\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Virtual Lessons<\/a><\/li>\n\t<\/ul>\n<\/li>\n\t<li class=\"menu-item menu-item-type-custom menu-item-object-custom menu-item-has-children menu-item-19890\"><a href=\"#\" class=\"elementor-sub-item elementor-item-anchor\" tabindex=\"-1\">Volume [Rotational]<\/a>\n\t<ul class=\"sub-menu elementor-nav-menu--dropdown\">\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-19931\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/integration-practice-problems-application-rotational-volumes\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Problem Set<\/a><\/li>\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-20160\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/virtual-lessons-integration-rotational-volume\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Virtual Lessons<\/a><\/li>\n\t<\/ul>\n<\/li>\n\t<li class=\"menu-item menu-item-type-custom menu-item-object-custom menu-item-has-children menu-item-19891\"><a href=\"#\" class=\"elementor-sub-item elementor-item-anchor\" tabindex=\"-1\">Surface Area<\/a>\n\t<ul class=\"sub-menu elementor-nav-menu--dropdown\">\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-19930\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/integration-practice-problems-application-surface-area\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Problem Set<\/a><\/li>\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-20165\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/virtual-lessons-integration-surface-area\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Virtual Lessons<\/a><\/li>\n\t<\/ul>\n<\/li>\n\t<li class=\"menu-item menu-item-type-custom menu-item-object-custom menu-item-has-children menu-item-19892\"><a href=\"#\" class=\"elementor-sub-item elementor-item-anchor\" tabindex=\"-1\">Arc Length<\/a>\n\t<ul class=\"sub-menu elementor-nav-menu--dropdown\">\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-20177\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/integration-practice-problems-arc-length\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Problem Set<\/a><\/li>\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-20176\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/virtual-lessons-integration-arc-length\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Virtual Lessons<\/a><\/li>\n\t<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n<li class=\"menu-item menu-item-type-custom menu-item-object-custom menu-item-has-children menu-item-19579\"><a href=\"#\" class=\"elementor-item elementor-item-anchor\" tabindex=\"-1\">Numerical Integration<\/a>\n<ul class=\"sub-menu elementor-nav-menu--dropdown\">\n\t<li class=\"menu-item menu-item-type-custom menu-item-object-custom menu-item-has-children menu-item-19894\"><a href=\"#\" class=\"elementor-sub-item elementor-item-anchor\" tabindex=\"-1\">Reimann Sums<\/a>\n\t<ul class=\"sub-menu elementor-nav-menu--dropdown\">\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-20110\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/integration-practice-problems-riemann-sums\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Problem Set<\/a><\/li>\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-20128\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/virtual-lessons-integration-riemann-sum\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Virtual Lessons<\/a><\/li>\n\t<\/ul>\n<\/li>\n\t<li class=\"menu-item menu-item-type-custom menu-item-object-custom menu-item-has-children menu-item-19895\"><a href=\"#\" class=\"elementor-sub-item elementor-item-anchor\" tabindex=\"-1\">Trapezoidal Rule<\/a>\n\t<ul class=\"sub-menu elementor-nav-menu--dropdown\">\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-20109\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/integration-practice-problems-trapezoidal\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Problem Set<\/a><\/li>\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-20138\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/virtual-lessons-integration-trapezoidal\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Virtual Lessons<\/a><\/li>\n\t<\/ul>\n<\/li>\n\t<li class=\"menu-item menu-item-type-custom menu-item-object-custom menu-item-has-children menu-item-19896\"><a href=\"#\" class=\"elementor-sub-item elementor-item-anchor\" tabindex=\"-1\">Simpson&#8217;s Rule<\/a>\n\t<ul class=\"sub-menu elementor-nav-menu--dropdown\">\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-20108\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/integration-practice-problems-simpsons\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Problem Set<\/a><\/li>\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-20137\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/virtual-lessons-integration-simpsons\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Virtual Lessons<\/a><\/li>\n\t<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n<\/ul>\t\t\t<\/nav>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t<section class=\"elementor-section elementor-top-section elementor-element elementor-element-51716149 elementor-section-stretched elementor-section-boxed elementor-section-height-default elementor-section-height-default\" data-id=\"51716149\" data-element_type=\"section\" data-settings=\"{&quot;stretch_section&quot;:&quot;section-stretched&quot;}\">\n\t\t\t\t\t\t<div class=\"elementor-container elementor-column-gap-default\">\n\t\t\t\t\t<div class=\"elementor-column elementor-col-100 elementor-top-column elementor-element elementor-element-70acdfc1\" data-id=\"70acdfc1\" data-element_type=\"column\">\n\t\t\t<div class=\"elementor-widget-wrap elementor-element-populated\">\n\t\t\t\t\t\t<div class=\"elementor-element elementor-element-525eca1f elementor-widget-divider--view-line elementor-widget elementor-widget-divider\" data-id=\"525eca1f\" data-element_type=\"widget\" data-widget_type=\"divider.default\">\n\t\t\t\t<div class=\"elementor-widget-container\">\n\t\t\t\t\t\t\t<div class=\"elementor-divider\">\n\t\t\t<span class=\"elementor-divider-separator\">\n\t\t\t\t\t\t<\/span>\n\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"elementor-element elementor-element-5d4b1c18 elementor-widget elementor-widget-text-editor\" data-id=\"5d4b1c18\" data-element_type=\"widget\" data-widget_type=\"text-editor.default\">\n\t\t\t\t<div class=\"elementor-widget-container\">\n\t\t\t\t\t\t\t\t\t<p style=\"text-align: justify;\"><span style=\"font-size: 18px;\">Integration by parts is an integration technique that involves separating the integrand into two parts. This technique is useful whenever a <strong>product<\/strong> appears in the integrand and the implement u-substitution is not feasible.<\/span><\/p><p style=\"text-align: center;\"><span style=\"font-size: 20px;\">\\(\\displaystyle\\int u\\,dv = uv\\,-\\!\\int v\\,du\\)<\/span><\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"elementor-element elementor-element-3c02949b elementor-widget-divider--view-line elementor-widget elementor-widget-divider\" data-id=\"3c02949b\" data-element_type=\"widget\" data-widget_type=\"divider.default\">\n\t\t\t\t<div class=\"elementor-widget-container\">\n\t\t\t\t\t\t\t<div class=\"elementor-divider\">\n\t\t\t<span class=\"elementor-divider-separator\">\n\t\t\t\t\t\t<\/span>\n\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t<section class=\"elementor-section elementor-top-section elementor-element elementor-element-6f62e114 elementor-section-boxed elementor-section-height-default elementor-section-height-default\" data-id=\"6f62e114\" data-element_type=\"section\">\n\t\t\t\t\t\t<div class=\"elementor-container elementor-column-gap-default\">\n\t\t\t\t\t<div class=\"elementor-column elementor-col-33 elementor-top-column elementor-element elementor-element-612534ae\" data-id=\"612534ae\" data-element_type=\"column\">\n\t\t\t<div class=\"elementor-widget-wrap\">\n\t\t\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t<div class=\"elementor-column elementor-col-33 elementor-top-column elementor-element elementor-element-3eaf959d\" data-id=\"3eaf959d\" data-element_type=\"column\">\n\t\t\t<div class=\"elementor-widget-wrap elementor-element-populated\">\n\t\t\t\t\t\t<div class=\"elementor-element elementor-element-35b3acae elementor-view-framed elementor-position-left elementor-shape-circle elementor-mobile-position-top elementor-vertical-align-top elementor-widget elementor-widget-icon-box\" data-id=\"35b3acae\" data-element_type=\"widget\" data-widget_type=\"icon-box.default\">\n\t\t\t\t<div class=\"elementor-widget-container\">\n\t\t\t\t\t\t\t<div class=\"elementor-icon-box-wrapper\">\n\n\t\t\t\t\t\t<div class=\"elementor-icon-box-icon\">\n\t\t\t\t<a href=\"https:\/\/onlinemathtutor.co\/index.php\/course-content\/calculus-resources\/integration\/problem-set-integration\/virtual-lessons-integration-ibp\/\" class=\"elementor-icon elementor-animation-grow\" tabindex=\"-1\" aria-label=\"Virtual Lessons\">\n\t\t\t\t<i aria-hidden=\"true\" class=\"fas fa-chalkboard-teacher\"><\/i>\t\t\t\t<\/a>\n\t\t\t<\/div>\n\t\t\t\n\t\t\t\t\t\t<div class=\"elementor-icon-box-content\">\n\n\t\t\t\t\t\t\t\t\t<h2 class=\"elementor-icon-box-title\">\n\t\t\t\t\t\t<a href=\"https:\/\/onlinemathtutor.co\/index.php\/course-content\/calculus-resources\/integration\/problem-set-integration\/virtual-lessons-integration-ibp\/\" >\n\t\t\t\t\t\t\tVirtual Lessons\t\t\t\t\t\t<\/a>\n\t\t\t\t\t<\/h2>\n\t\t\t\t\n\t\t\t\t\t\t\t\t\t<p class=\"elementor-icon-box-description\">\n\t\t\t\t\t\tNeed some additional help understanding how to apply this integration technique? Click Here to visit the virtual lesson section. \t\t\t\t\t<\/p>\n\t\t\t\t\n\t\t\t<\/div>\n\t\t\t\n\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t<div class=\"elementor-column elementor-col-33 elementor-top-column elementor-element elementor-element-7471220\" data-id=\"7471220\" data-element_type=\"column\">\n\t\t\t<div class=\"elementor-widget-wrap\">\n\t\t\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t<section class=\"elementor-section elementor-top-section elementor-element elementor-element-c281c6a elementor-section-boxed elementor-section-height-default elementor-section-height-default\" data-id=\"c281c6a\" data-element_type=\"section\">\n\t\t\t\t\t\t<div class=\"elementor-container elementor-column-gap-default\">\n\t\t\t\t\t<div class=\"elementor-column elementor-col-100 elementor-top-column elementor-element elementor-element-34ebefad\" data-id=\"34ebefad\" data-element_type=\"column\">\n\t\t\t<div class=\"elementor-widget-wrap elementor-element-populated\">\n\t\t\t\t\t\t<div class=\"elementor-element elementor-element-6f66d3b8 elementor-widget-divider--view-line elementor-widget elementor-widget-divider\" data-id=\"6f66d3b8\" data-element_type=\"widget\" data-widget_type=\"divider.default\">\n\t\t\t\t<div class=\"elementor-widget-container\">\n\t\t\t\t\t\t\t<div class=\"elementor-divider\">\n\t\t\t<span class=\"elementor-divider-separator\">\n\t\t\t\t\t\t<\/span>\n\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"elementor-element elementor-element-26316a48 elementor-widget elementor-widget-spacer\" data-id=\"26316a48\" data-element_type=\"widget\" data-widget_type=\"spacer.default\">\n\t\t\t\t<div class=\"elementor-widget-container\">\n\t\t\t\t\t\t\t<div class=\"elementor-spacer\">\n\t\t\t<div class=\"elementor-spacer-inner\"><\/div>\n\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t<section class=\"elementor-section elementor-top-section elementor-element elementor-element-12d58213 elementor-section-boxed elementor-section-height-default elementor-section-height-default\" data-id=\"12d58213\" data-element_type=\"section\">\n\t\t\t\t\t\t<div class=\"elementor-container elementor-column-gap-default\">\n\t\t\t\t\t<div class=\"elementor-column elementor-col-100 elementor-top-column elementor-element elementor-element-51dcc7ca\" data-id=\"51dcc7ca\" data-element_type=\"column\">\n\t\t\t<div class=\"elementor-widget-wrap elementor-element-populated\">\n\t\t\t\t\t\t<div class=\"elementor-element elementor-element-fe0e51c elementor-widget elementor-widget-heading\" data-id=\"fe0e51c\" data-element_type=\"widget\" data-widget_type=\"heading.default\">\n\t\t\t\t<div class=\"elementor-widget-container\">\n\t\t\t\t\t<h2 class=\"elementor-heading-title elementor-size-large\">Practice Problems<\/h2>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"elementor-element elementor-element-5ab74305 elementor-widget elementor-widget-text-editor\" data-id=\"5ab74305\" data-element_type=\"widget\" data-widget_type=\"text-editor.default\">\n\t\t\t\t<div class=\"elementor-widget-container\">\n\t\t\t\t\t\t\t\t\t<p style=\"text-align: justify;\"><span style=\"font-size: 18px;\">Answer the following questions by using the integration technique known as <strong>integration by parts (IBP)<\/strong>.\u00a0<\/span><\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t<section class=\"elementor-section elementor-top-section elementor-element elementor-element-5a340b6 elementor-section-boxed elementor-section-height-default elementor-section-height-default\" data-id=\"5a340b6\" data-element_type=\"section\">\n\t\t\t\t\t\t<div class=\"elementor-container elementor-column-gap-default\">\n\t\t\t\t\t<div class=\"elementor-column elementor-col-33 elementor-top-column elementor-element elementor-element-4f74bdce\" data-id=\"4f74bdce\" data-element_type=\"column\">\n\t\t\t<div class=\"elementor-widget-wrap elementor-element-populated\">\n\t\t\t\t\t\t<div class=\"elementor-element elementor-element-1cb47e95 elementor-widget elementor-widget-text-editor\" data-id=\"1cb47e95\" data-element_type=\"widget\" data-widget_type=\"text-editor.default\">\n\t\t\t\t<div class=\"elementor-widget-container\">\n\t\t\t\t\t\t\t\t\t<ol start=\"1\">\n \t<li style=\"margin-bottom: 30px;\">\\(\\displaystyle \\int\u00a0 x\\texttt{sin}(2x) dx \\)<\/li>\n \t<li style=\"margin-bottom: 30px;\">\\(\\displaystyle \\int\u00a0x^2\\texttt{sin}(2x) dx \\)<\/li>\n \t<li style=\"margin-bottom: 30px;\">\\(\\displaystyle \\int \\texttt{arctan}(2x+1) dx \\)<\/li>\n \t<li style=\"margin-bottom: 30px;\">\\(\\displaystyle \\int \\texttt{cos}(\\sqrt{x}) dx \\)<\/li>\n<\/ol>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t<div class=\"elementor-column elementor-col-33 elementor-top-column elementor-element elementor-element-3900047a\" data-id=\"3900047a\" data-element_type=\"column\">\n\t\t\t<div class=\"elementor-widget-wrap elementor-element-populated\">\n\t\t\t\t\t\t<div class=\"elementor-element elementor-element-45269ad elementor-widget elementor-widget-text-editor\" data-id=\"45269ad\" data-element_type=\"widget\" data-widget_type=\"text-editor.default\">\n\t\t\t\t<div class=\"elementor-widget-container\">\n\t\t\t\t\t\t\t\t\t<ol start=\"5\">\n \t<li style=\"margin-bottom: 30px;\">\\(\\displaystyle \\int \\texttt{e}^{\\sqrt{x}} dx \\)<\/li>\n \t<li style=\"margin-bottom: 30px;\">\\(\\displaystyle \\int \\texttt{sin}(\\texttt{ln}(x)) dx \\)<\/li>\n \t<li style=\"margin-bottom: 30px;\">\\(\\displaystyle \\int \\texttt{e}^{2x}\\texttt{cos}(5x) dx \\)<\/li>\n \t<li style=\"margin-bottom: 30px;\">\\(\\displaystyle \\int \\texttt{e}^{\\alpha x}\\texttt{sin}(\\beta x)dx \\)<\/li>\n<\/ol>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t<div class=\"elementor-column elementor-col-33 elementor-top-column elementor-element elementor-element-39b19e29\" data-id=\"39b19e29\" data-element_type=\"column\">\n\t\t\t<div class=\"elementor-widget-wrap elementor-element-populated\">\n\t\t\t\t\t\t<div class=\"elementor-element elementor-element-7c3f546d elementor-widget elementor-widget-text-editor\" data-id=\"7c3f546d\" data-element_type=\"widget\" data-widget_type=\"text-editor.default\">\n\t\t\t\t<div class=\"elementor-widget-container\">\n\t\t\t\t\t\t\t\t\t<ol start=\"9\">\n \t<li style=\"margin-bottom: 30px;\">\\(\\displaystyle \\int\u00a0 \\texttt{e}^{\\alpha x}\\texttt{cos}(\\beta x) dx \\)<\/li>\n<\/ol>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t<section class=\"elementor-section elementor-top-section elementor-element elementor-element-5bbd798 elementor-section-boxed elementor-section-height-default elementor-section-height-default\" data-id=\"5bbd798\" data-element_type=\"section\">\n\t\t\t\t\t\t<div class=\"elementor-container elementor-column-gap-default\">\n\t\t\t\t\t<div class=\"elementor-column elementor-col-100 elementor-top-column elementor-element elementor-element-10d97dfe\" data-id=\"10d97dfe\" data-element_type=\"column\">\n\t\t\t<div class=\"elementor-widget-wrap elementor-element-populated\">\n\t\t\t\t\t\t<div class=\"elementor-element elementor-element-37c3e202 elementor-align-left elementor-icon-list--layout-traditional elementor-list-item-link-full_width elementor-widget elementor-widget-icon-list\" data-id=\"37c3e202\" data-element_type=\"widget\" data-widget_type=\"icon-list.default\">\n\t\t\t\t<div class=\"elementor-widget-container\">\n\t\t\t\t\t\t\t<ul class=\"elementor-icon-list-items\">\n\t\t\t\t\t\t\t<li class=\"elementor-icon-list-item\">\n\t\t\t\t\t\t\t\t\t\t\t<a href=\"https:\/\/onlinemathtutor.co\/index.php\/problem-set-integration\/\">\n\n\t\t\t\t\t\t\t\t\t\t\t\t<span class=\"elementor-icon-list-icon\">\n\t\t\t\t\t\t\t<i aria-hidden=\"true\" class=\"fas fa-arrow-circle-right\"><\/i>\t\t\t\t\t\t<\/span>\n\t\t\t\t\t\t\t\t\t\t<span class=\"elementor-icon-list-text\">Back To Main Page: Integration Problem Sets<\/span>\n\t\t\t\t\t\t\t\t\t\t\t<\/a>\n\t\t\t\t\t\t\t\t\t<\/li>\n\t\t\t\t\t\t<\/ul>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"elementor-element elementor-element-7c81c5b3 elementor-widget elementor-widget-spacer\" data-id=\"7c81c5b3\" data-element_type=\"widget\" data-widget_type=\"spacer.default\">\n\t\t\t\t<div class=\"elementor-widget-container\">\n\t\t\t\t\t\t\t<div class=\"elementor-spacer\">\n\t\t\t<div class=\"elementor-spacer-inner\"><\/div>\n\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t<section class=\"elementor-section elementor-top-section elementor-element elementor-element-c68e88b elementor-section-boxed elementor-section-height-default elementor-section-height-default\" data-id=\"c68e88b\" data-element_type=\"section\">\n\t\t\t\t\t\t<div class=\"elementor-container elementor-column-gap-default\">\n\t\t\t\t\t<div class=\"elementor-column elementor-col-100 elementor-top-column elementor-element elementor-element-26d14011\" data-id=\"26d14011\" data-element_type=\"column\">\n\t\t\t<div class=\"elementor-widget-wrap elementor-element-populated\">\n\t\t\t\t\t\t<div class=\"elementor-element elementor-element-3e74e8b4 elementor-widget elementor-widget-heading\" data-id=\"3e74e8b4\" data-element_type=\"widget\" data-widget_type=\"heading.default\">\n\t\t\t\t<div class=\"elementor-widget-container\">\n\t\t\t\t\t<h2 class=\"elementor-heading-title elementor-size-large\">Suggestive Solution Guide<\/h2>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t<section class=\"elementor-section elementor-top-section elementor-element elementor-element-27e7c477 elementor-section-boxed elementor-section-height-default elementor-section-height-default\" data-id=\"27e7c477\" data-element_type=\"section\">\n\t\t\t\t\t\t<div class=\"elementor-container elementor-column-gap-default\">\n\t\t\t\t\t<div class=\"elementor-column elementor-col-100 elementor-top-column elementor-element elementor-element-fcf1e30\" data-id=\"fcf1e30\" data-element_type=\"column\">\n\t\t\t<div class=\"elementor-widget-wrap elementor-element-populated\">\n\t\t\t\t\t\t<div class=\"elementor-element elementor-element-e4f9fa1 elementor-widget elementor-widget-toggle\" data-id=\"e4f9fa1\" data-element_type=\"widget\" data-widget_type=\"toggle.default\">\n\t\t\t\t<div class=\"elementor-widget-container\">\n\t\t\t\t\t\t\t<div class=\"elementor-toggle\">\n\t\t\t\t\t\t\t<div class=\"elementor-toggle-item\">\n\t\t\t\t\t<h2 id=\"elementor-tab-title-2401\" class=\"elementor-tab-title\" data-tab=\"1\" role=\"button\" aria-controls=\"elementor-tab-content-2401\" aria-expanded=\"false\">\n\t\t\t\t\t\t\t\t\t\t\t\t<span class=\"elementor-toggle-icon elementor-toggle-icon-left\" aria-hidden=\"true\">\n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t<span class=\"elementor-toggle-icon-closed\"><i class=\"fas fa-toggle-off\"><\/i><\/span>\n\t\t\t\t\t\t\t\t<span class=\"elementor-toggle-icon-opened\"><i class=\"elementor-toggle-icon-opened fas fa-toggle-on\"><\/i><\/span>\n\t\t\t\t\t\t\t\t\t\t\t\t\t<\/span>\n\t\t\t\t\t\t\t\t\t\t\t\t<a class=\"elementor-toggle-title\" tabindex=\"0\">Problem 1: Solution<\/a>\n\t\t\t\t\t<\/h2>\n\n\t\t\t\t\t<div id=\"elementor-tab-content-2401\" class=\"elementor-tab-content elementor-clearfix\" data-tab=\"1\" role=\"region\" aria-labelledby=\"elementor-tab-title-2401\"><p><span style=\"font-size: 18.6667px;\">Let \\(u=x\\), so \\(du=dx\\); and \\(dv=\\texttt{sin}(2x)dx\\), so\u00a0<\/span><span style=\"font-size: 18.6667px;\">\\(v=\\int\\texttt{sin}(2x)dx=-\\frac{1}{2}\\texttt{cos}(2x)\\)<\/span><span style=\"font-size: 18.6667px;\">.<\/span><span style=\"font-size: 18.6667px;\">\u00a0<\/span><span style=\"font-size: 18.6667px;\">Substituting these values into \\(uv\\:- \\int vdu\\) leads to the following:<\/span><\/p>\\(\\normalsize<br \/>\\begin{align*}\\displaystyle<br \/>\\int x\\texttt{sin}(2x) dx<br \/>&amp;= \\color{red}{\\underbrace{\\color{black}{x}}_{\\large u}} \\color{red}{\\underbrace{ \\color{black}{\\left( \\frac{-1}{2}\\texttt{cos}(2x) \\right)}}_{\\large v}}\\:- \\int\\ \\color{red}{\\underbrace{\\color{black}{\\left(\\frac{-1}{2}\\texttt{cos}(2x)\\right)}}_{\\large v}} \\color{red}{\\underbrace{\\color{black}{dx}}_{\\large du}}<br \/>&amp; \\color{gray}{\\text{[Apply the substitution]}} \\\\[5pt]<br \/>&amp;= \\frac{-x}{2}\\texttt{cos}(2x)\\:+ \\frac{1}{2}\\int \\texttt{cos}(2x)dx<br \/>&amp; \\color{gray}{\\text{[Rearrange the expression]}} \\\\[5pt]<br \/>&amp;= \\color{green}{\\frac{-x}{2}\\texttt{cos}(2x)\\:+ \\frac{1}{4}\\texttt{sin}(2x)\\:+ C}<br \/>&amp; \\color{gray}{\\text{[Integrate using a quick u-sub]}} \\\\[5pt]<br \/>\\end{align*}<br \/>\\)<p>\u00a0<\/p><\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t\t\t\t\t<script type=\"application\/ld+json\">{\"@context\":\"https:\\\/\\\/schema.org\",\"@type\":\"FAQPage\",\"mainEntity\":[{\"@type\":\"Question\",\"name\":\"Problem 1: Solution\",\"acceptedAnswer\":{\"@type\":\"Answer\",\"text\":\"<p><span style=\\\"font-size: 18.6667px;\\\">Let \\\\(u=x\\\\), so \\\\(du=dx\\\\); and \\\\(dv=\\\\texttt{sin}(2x)dx\\\\), so\\u00a0<\\\/span><span style=\\\"font-size: 18.6667px;\\\">\\\\(v=\\\\int\\\\texttt{sin}(2x)dx=-\\\\frac{1}{2}\\\\texttt{cos}(2x)\\\\)<\\\/span><span style=\\\"font-size: 18.6667px;\\\">.<\\\/span><span style=\\\"font-size: 18.6667px;\\\">\\u00a0<\\\/span><span style=\\\"font-size: 18.6667px;\\\">Substituting these values into \\\\(uv\\\\:- \\\\int vdu\\\\) leads to the following:<\\\/span><\\\/p>\\\\(\\\\normalsize<br \\\/>\\\\begin{align*}\\\\displaystyle<br \\\/>\\\\int x\\\\texttt{sin}(2x) dx<br \\\/>&amp;= \\\\color{red}{\\\\underbrace{\\\\color{black}{x}}_{\\\\large u}} \\\\color{red}{\\\\underbrace{ \\\\color{black}{\\\\left( \\\\frac{-1}{2}\\\\texttt{cos}(2x) \\\\right)}}_{\\\\large v}}\\\\:- \\\\int\\\\ \\\\color{red}{\\\\underbrace{\\\\color{black}{\\\\left(\\\\frac{-1}{2}\\\\texttt{cos}(2x)\\\\right)}}_{\\\\large v}} \\\\color{red}{\\\\underbrace{\\\\color{black}{dx}}_{\\\\large du}}<br \\\/>&amp; \\\\color{gray}{\\\\text{[Apply the substitution]}} \\\\\\\\[5pt]<br \\\/>&amp;= \\\\frac{-x}{2}\\\\texttt{cos}(2x)\\\\:+ \\\\frac{1}{2}\\\\int \\\\texttt{cos}(2x)dx<br \\\/>&amp; \\\\color{gray}{\\\\text{[Rearrange the expression]}} \\\\\\\\[5pt]<br \\\/>&amp;= \\\\color{green}{\\\\frac{-x}{2}\\\\texttt{cos}(2x)\\\\:+ \\\\frac{1}{4}\\\\texttt{sin}(2x)\\\\:+ C}<br \\\/>&amp; \\\\color{gray}{\\\\text{[Integrate using a quick u-sub]}} \\\\\\\\[5pt]<br \\\/>\\\\end{align*}<br \\\/>\\\\)<p>\\u00a0<\\\/p>\"}}]}<\/script>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t<section class=\"elementor-section elementor-top-section elementor-element elementor-element-3dcddb98 elementor-section-boxed elementor-section-height-default elementor-section-height-default\" data-id=\"3dcddb98\" data-element_type=\"section\">\n\t\t\t\t\t\t<div class=\"elementor-container elementor-column-gap-default\">\n\t\t\t\t\t<div class=\"elementor-column elementor-col-100 elementor-top-column elementor-element elementor-element-5842b8dc\" data-id=\"5842b8dc\" data-element_type=\"column\">\n\t\t\t<div class=\"elementor-widget-wrap elementor-element-populated\">\n\t\t\t\t\t\t<div class=\"elementor-element elementor-element-4fce3554 elementor-widget elementor-widget-toggle\" data-id=\"4fce3554\" data-element_type=\"widget\" data-widget_type=\"toggle.default\">\n\t\t\t\t<div class=\"elementor-widget-container\">\n\t\t\t\t\t\t\t<div class=\"elementor-toggle\">\n\t\t\t\t\t\t\t<div class=\"elementor-toggle-item\">\n\t\t\t\t\t<h2 id=\"elementor-tab-title-1331\" class=\"elementor-tab-title\" data-tab=\"1\" role=\"button\" aria-controls=\"elementor-tab-content-1331\" aria-expanded=\"false\">\n\t\t\t\t\t\t\t\t\t\t\t\t<span class=\"elementor-toggle-icon elementor-toggle-icon-left\" aria-hidden=\"true\">\n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t<span class=\"elementor-toggle-icon-closed\"><i class=\"fas fa-toggle-off\"><\/i><\/span>\n\t\t\t\t\t\t\t\t<span class=\"elementor-toggle-icon-opened\"><i class=\"elementor-toggle-icon-opened fas fa-toggle-on\"><\/i><\/span>\n\t\t\t\t\t\t\t\t\t\t\t\t\t<\/span>\n\t\t\t\t\t\t\t\t\t\t\t\t<a class=\"elementor-toggle-title\" tabindex=\"0\">Problem 2: Solution<\/a>\n\t\t\t\t\t<\/h2>\n\n\t\t\t\t\t<div id=\"elementor-tab-content-1331\" class=\"elementor-tab-content elementor-clearfix\" data-tab=\"1\" role=\"region\" aria-labelledby=\"elementor-tab-title-1331\"><p><span style=\"font-size: 18.6667px;\">Let \\(u=x^2\\), so \\(du=2xdx\\); and \\(dv=\\texttt{sin}(2x)dx\\), so \\(v = \\int \\texttt{sin}(2x)dx = -\\frac{1}{2}\\texttt{cos}(2x)\\).<\/span><span style=\"font-size: 18.6667px;\">\u00a0<\/span><span style=\"font-size: 18.6667px;\">Substituting these values into \\(uv\\:- \\int vdu\\) leads to the following:<\/span><\/p>\\(\\normalsize<br \/>\\begin{align*}\\displaystyle<br \/>\\int x^2\\texttt{sin}(2x)dx<br \/>&amp;= \\color{red}{\\underbrace{\\color{black}{x^2}}_{\\large u}} \\color{red}{\\underbrace{\\color{black}{\\left( \\frac{-1}{2}\\texttt{cos}(2x) \\right)}}_{\\large v}}\\:- \\int \\color{red}{\\underbrace{\\color{black}{\\left(\\frac{-1}{2}\\texttt{cos}(2x)\\right)}}_{\\large v}} \\color{red}{\\underbrace{\\color{black}{(2xdx)}}_{\\large du}}<br \/>&amp; \\color{gray}{\\text{[Apply the substitution]}} \\\\[5pt]<br \/>&amp;= -\\frac{x^2}{2}cos(2x)\\:+ \\int x\\texttt{cos}(2x)dx<br \/>&amp; \\color{gray}{\\text{[Rearrange the expression]}} \\\\[5pt]<br \/>\\end{align*}<br \/>\\)<p><span style=\"font-size: 18.6667px;\">\u00a0<\/span><\/p><p><span style=\"font-size: 18.6667px;\">Apply IBP once again. Let \\(u=x\\), so \\(du=dx\\); and \\(dv=\\texttt{cos}(2x)dx\\), so \\(v = \\int \\texttt{cos}(2x)dx = \\frac{1}{2}\\texttt{sin}(2x)\\).<\/span><span style=\"font-size: 18.6667px;\">\u00a0<\/span><span style=\"font-size: 18.6667px;\">Substituting these values into \\(uv\\:- \\int vdu\\) leads to the following:<\/span><\/p>\\(\\normalsize<br \/>\\begin{align*}\\displaystyle<br \/>-\\frac{x^2}{2}\\texttt{cos}(2x)\\:+ \\int x\\texttt{cos}(2x)dx<br \/>&amp;= -\\frac{x^2}{2}\\texttt{cos}(2x)\\:+ \\left(\\color{red}{\\underbrace{\\color{black}{x}}_{\\large u}} \\color{red}{\\underbrace{\\color{black}{\\left( \\frac{1}{2}\\texttt{sin}(2x) \\right)}}_{\\large v}}\\:- \\int \\color{red}{\\underbrace{\\color{black}{\\left(\\frac{1}{2}\\texttt{sin}(2x)\\right)}}_{\\large v}} \\color{red}{\\underbrace{\\color{black}{(dx)}}_{\\large du}}\\right)<br \/>&amp; \\color{gray}{\\text{[Apply the substitution]}} \\\\[5pt]<br \/>&amp;= -\\frac{x^2}{2}\\texttt{cos}(2x)\\:+ \\frac{x}{2}\\texttt{sin}(2x)\\:- \\frac{1}{2}\\int \\texttt{sin}(2x)dx<br \/>&amp; \\color{gray}{\\text{[Rearrange the expression]}} \\\\[5pt]<br \/>&amp;= -\\frac{x^2}{2}\\texttt{cos}(2x)\\:+ \\frac{x}{2}\\texttt{sin}(2x)\\:- \\left(\\frac{1}{2}\\right) \\left(\\frac{-1}{2}\\texttt{cos}(2x)\\right)\\:+ C<br \/>&amp; \\color{gray}{\\text{[Integrate using a quick u-sub]}} \\\\[5pt]<br \/>&amp;= \\color{green}{-\\frac{x^2}{2}\\texttt{cos}(2x)\\:+ \\frac{x}{2}\\texttt{sin}(2x)\\:+ \\frac{1}{4}\\texttt{cos}(2x)\\:+ C}<br \/>&amp; \\color{gray}{\\text{[Rearrange the expression]}} \\\\[5pt]<br \/>\\end{align*}<br \/>\\)<p>\u00a0<\/p><\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t\t\t\t\t<script type=\"application\/ld+json\">{\"@context\":\"https:\\\/\\\/schema.org\",\"@type\":\"FAQPage\",\"mainEntity\":[{\"@type\":\"Question\",\"name\":\"Problem 2: Solution\",\"acceptedAnswer\":{\"@type\":\"Answer\",\"text\":\"<p><span style=\\\"font-size: 18.6667px;\\\">Let \\\\(u=x^2\\\\), so \\\\(du=2xdx\\\\); and \\\\(dv=\\\\texttt{sin}(2x)dx\\\\), so \\\\(v = \\\\int \\\\texttt{sin}(2x)dx = -\\\\frac{1}{2}\\\\texttt{cos}(2x)\\\\).<\\\/span><span style=\\\"font-size: 18.6667px;\\\">\\u00a0<\\\/span><span style=\\\"font-size: 18.6667px;\\\">Substituting these values into \\\\(uv\\\\:- \\\\int vdu\\\\) leads to the following:<\\\/span><\\\/p>\\\\(\\\\normalsize<br \\\/>\\\\begin{align*}\\\\displaystyle<br \\\/>\\\\int x^2\\\\texttt{sin}(2x)dx<br \\\/>&amp;= \\\\color{red}{\\\\underbrace{\\\\color{black}{x^2}}_{\\\\large u}} \\\\color{red}{\\\\underbrace{\\\\color{black}{\\\\left( \\\\frac{-1}{2}\\\\texttt{cos}(2x) \\\\right)}}_{\\\\large v}}\\\\:- \\\\int \\\\color{red}{\\\\underbrace{\\\\color{black}{\\\\left(\\\\frac{-1}{2}\\\\texttt{cos}(2x)\\\\right)}}_{\\\\large v}} \\\\color{red}{\\\\underbrace{\\\\color{black}{(2xdx)}}_{\\\\large du}}<br \\\/>&amp; \\\\color{gray}{\\\\text{[Apply the substitution]}} \\\\\\\\[5pt]<br \\\/>&amp;= -\\\\frac{x^2}{2}cos(2x)\\\\:+ \\\\int x\\\\texttt{cos}(2x)dx<br \\\/>&amp; \\\\color{gray}{\\\\text{[Rearrange the expression]}} \\\\\\\\[5pt]<br \\\/>\\\\end{align*}<br \\\/>\\\\)<p><span style=\\\"font-size: 18.6667px;\\\">\\u00a0<\\\/span><\\\/p><p><span style=\\\"font-size: 18.6667px;\\\">Apply IBP once again. Let \\\\(u=x\\\\), so \\\\(du=dx\\\\); and \\\\(dv=\\\\texttt{cos}(2x)dx\\\\), so \\\\(v = \\\\int \\\\texttt{cos}(2x)dx = \\\\frac{1}{2}\\\\texttt{sin}(2x)\\\\).<\\\/span><span style=\\\"font-size: 18.6667px;\\\">\\u00a0<\\\/span><span style=\\\"font-size: 18.6667px;\\\">Substituting these values into \\\\(uv\\\\:- \\\\int vdu\\\\) leads to the following:<\\\/span><\\\/p>\\\\(\\\\normalsize<br \\\/>\\\\begin{align*}\\\\displaystyle<br \\\/>-\\\\frac{x^2}{2}\\\\texttt{cos}(2x)\\\\:+ \\\\int x\\\\texttt{cos}(2x)dx<br \\\/>&amp;= -\\\\frac{x^2}{2}\\\\texttt{cos}(2x)\\\\:+ \\\\left(\\\\color{red}{\\\\underbrace{\\\\color{black}{x}}_{\\\\large u}} \\\\color{red}{\\\\underbrace{\\\\color{black}{\\\\left( \\\\frac{1}{2}\\\\texttt{sin}(2x) \\\\right)}}_{\\\\large v}}\\\\:- \\\\int \\\\color{red}{\\\\underbrace{\\\\color{black}{\\\\left(\\\\frac{1}{2}\\\\texttt{sin}(2x)\\\\right)}}_{\\\\large v}} \\\\color{red}{\\\\underbrace{\\\\color{black}{(dx)}}_{\\\\large du}}\\\\right)<br \\\/>&amp; \\\\color{gray}{\\\\text{[Apply the substitution]}} \\\\\\\\[5pt]<br \\\/>&amp;= -\\\\frac{x^2}{2}\\\\texttt{cos}(2x)\\\\:+ \\\\frac{x}{2}\\\\texttt{sin}(2x)\\\\:- \\\\frac{1}{2}\\\\int \\\\texttt{sin}(2x)dx<br \\\/>&amp; \\\\color{gray}{\\\\text{[Rearrange the expression]}} \\\\\\\\[5pt]<br \\\/>&amp;= -\\\\frac{x^2}{2}\\\\texttt{cos}(2x)\\\\:+ \\\\frac{x}{2}\\\\texttt{sin}(2x)\\\\:- \\\\left(\\\\frac{1}{2}\\\\right) \\\\left(\\\\frac{-1}{2}\\\\texttt{cos}(2x)\\\\right)\\\\:+ C<br \\\/>&amp; \\\\color{gray}{\\\\text{[Integrate using a quick u-sub]}} \\\\\\\\[5pt]<br \\\/>&amp;= \\\\color{green}{-\\\\frac{x^2}{2}\\\\texttt{cos}(2x)\\\\:+ \\\\frac{x}{2}\\\\texttt{sin}(2x)\\\\:+ \\\\frac{1}{4}\\\\texttt{cos}(2x)\\\\:+ C}<br \\\/>&amp; \\\\color{gray}{\\\\text{[Rearrange the expression]}} \\\\\\\\[5pt]<br \\\/>\\\\end{align*}<br \\\/>\\\\)<p>\\u00a0<\\\/p>\"}}]}<\/script>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t<section class=\"elementor-section elementor-top-section elementor-element elementor-element-17c72b1e elementor-section-boxed elementor-section-height-default elementor-section-height-default\" data-id=\"17c72b1e\" data-element_type=\"section\">\n\t\t\t\t\t\t<div class=\"elementor-container elementor-column-gap-default\">\n\t\t\t\t\t<div class=\"elementor-column elementor-col-100 elementor-top-column elementor-element elementor-element-7942ea93\" data-id=\"7942ea93\" data-element_type=\"column\">\n\t\t\t<div class=\"elementor-widget-wrap elementor-element-populated\">\n\t\t\t\t\t\t<div class=\"elementor-element elementor-element-14b5fee4 elementor-widget elementor-widget-toggle\" data-id=\"14b5fee4\" data-element_type=\"widget\" data-widget_type=\"toggle.default\">\n\t\t\t\t<div class=\"elementor-widget-container\">\n\t\t\t\t\t\t\t<div class=\"elementor-toggle\">\n\t\t\t\t\t\t\t<div class=\"elementor-toggle-item\">\n\t\t\t\t\t<h2 id=\"elementor-tab-title-3471\" class=\"elementor-tab-title\" data-tab=\"1\" role=\"button\" aria-controls=\"elementor-tab-content-3471\" aria-expanded=\"false\">\n\t\t\t\t\t\t\t\t\t\t\t\t<span class=\"elementor-toggle-icon elementor-toggle-icon-left\" aria-hidden=\"true\">\n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t<span class=\"elementor-toggle-icon-closed\"><i class=\"fas fa-toggle-off\"><\/i><\/span>\n\t\t\t\t\t\t\t\t<span class=\"elementor-toggle-icon-opened\"><i class=\"elementor-toggle-icon-opened fas fa-toggle-on\"><\/i><\/span>\n\t\t\t\t\t\t\t\t\t\t\t\t\t<\/span>\n\t\t\t\t\t\t\t\t\t\t\t\t<a class=\"elementor-toggle-title\" tabindex=\"0\">Problem 3: Solution<\/a>\n\t\t\t\t\t<\/h2>\n\n\t\t\t\t\t<div id=\"elementor-tab-content-3471\" class=\"elementor-tab-content elementor-clearfix\" data-tab=\"1\" role=\"region\" aria-labelledby=\"elementor-tab-title-3471\"><p><span style=\"font-size: 18.6667px;\">Let \\(u=\\texttt{arctan}(2x+1)\\), so \\(du=\\frac{2}{(2x+1)^2+1}dx\\); and \\(dv=dx\\), so \\(v=x\\).<\/span><span style=\"font-size: 18.6667px;\">\u00a0<\/span><span style=\"font-size: 18.6667px;\">Substituting these values into \\(uv\\:- \\int vdu\\) leads to the following:<\/span><\/p>\\(\\normalsize<br \/>\\begin{align*}\\displaystyle<br \/>\\int \\texttt{arctan}(2x+1)dx<br \/>&amp;= \\color{red}{\\underbrace{\\color{black}{\\texttt{arctan}(2x+1)}}_{\\large u}} \\color{red}{\\underbrace{\\color{black}{x}}_{\\large v}}\\:- \\int \\color{red}{\\underbrace{\\color{black}{x}}_{\\large v}} \\color{red}{\\underbrace{\\color{black}{\\left(\\frac{2}{(2x+1)^2+1}\\right)}}_{\\large du}}dx<br \/>&amp; \\color{gray}{\\text{[Apply the substitution]}} \\\\[5pt]<br \/>&amp;= x\\texttt{arctan}(2x+1)\\:- 2\\int \\frac{x}{(2x+1)^2+1}dx<br \/>&amp; \\color{gray}{\\text{[Rearrange the expression]}} \\\\[5pt]<br \/>\\end{align*}<br \/>\\)<p><span style=\"font-size: 18.6667px;\">\u00a0<\/span><\/p><p><span style=\"font-size: 18.6667px;\">Use U-Sub with the given integrand. Let \\(u=2x+1\\), so \\(du=2dx\\); and \\(x=\\frac{u-1}{2}\\). While working the only integral left, it&#8217;s fine to have \\(u\\) and \\(x\\) at the same time given that the integrand is expressed in regards of just one variable. Substituting accordingly yields:<\/span><\/p>\\(\\normalsize<br \/>\\begin{align*}\\displaystyle<br \/>x\\texttt{arctan}(2x+1)\\:- 2\\int \\frac{x}{(2x+1)^2+1}dx<br \/>&amp;= x\\texttt{arctan}(2x+1)\\:- 2\\int \\left( \\frac{\\color{red}{\\left( \\frac{u-1}{2} \\right)}}{\\color{red}{u^2}+1} \\right) \\color{red}{\\left( \\frac{du}{2} \\right)}<br \/>&amp; \\color{gray}{\\text{[Apply the substitution]}} \\\\[5pt]<br \/>&amp;= x\\texttt{arctan}(2x+1)\\:- \\frac{1}{2}\\int\\frac{u-1}{u^2+1}du<br \/>&amp; \\color{gray}{\\text{[Rearrange the expression]}} \\\\[5pt]<br \/>&amp;= x\\texttt{arctan}(2x+1)\\:- \\frac{1}{2}\\left( \\int\\frac{u}{u^2+1}du\\:-\\int\\frac{1}{u^2+1}du \\right)<br \/>&amp; \\color{gray}{\\text{[Split the integrals strategically]}} \\\\[5pt]<br \/>&amp;= x\\texttt{arctan}(2x+1)\\:- \\frac{1}{2}\\int\\frac{u}{u^2+1}du\\:+\\frac{1}{2}\\int\\frac{1}{u^2+1}du<br \/>&amp; \\color{gray}{\\text{[Distribute the fraction]}} \\\\[5pt]<br \/>&amp;= x\\texttt{arctan}(2x+1)\\:+ \\frac{1}{2}\\texttt{arctan}(u)\\:- \\frac{1}{2}\\int\\frac{u}{u^2+1}du<br \/>&amp; \\color{gray}{\\text{[Integrate the second integral]}} \\\\[5pt]<br \/>&amp;= x\\texttt{arctan}(2x+1)\\:+ \\frac{1}{2}\\texttt{arctan}(2x+1)\\:- \\frac{1}{2}\\int\\frac{u}{u^2+1}du<br \/>&amp; \\color{gray}{\\text{[Substitute x back to the recently integrated term]}} \\\\[5pt]<br \/>&amp;= \\texttt{arctan}(2x+1) \\left[ x+\\frac{1}{2} \\right] \\:- \\frac{1}{2}\\int\\frac{u}{u^2+1}du<br \/>&amp; \\color{gray}{\\text{[Factor the arctan]}} \\\\[5pt]<br \/>\\end{align*}<br \/>\\)<p><span style=\"font-size: 18.6667px;\">\u00a0<\/span><\/p><p><span style=\"font-size: 18.6667px;\">Use U-Sub with the given integrand. Let \\(v=u^2\\), so \\(dv=2udu\\)<\/span><span style=\"font-size: 18.6667px;\">. Substituting accordingly yields:<\/span><\/p>\\(\\normalsize<br \/>\\begin{align*}\\displaystyle<br \/>\\texttt{arctan}(2x+1) \\left[ x+\\frac{1}{2} \\right] \\:- \\frac{1}{2}\\int\\frac{u}{u^2+1}du<br \/>&amp;= \\texttt{arctan}(2x+1) \\left[ x+\\frac{1}{2} \\right] \\:- \\frac{1}{2}\\int\\frac{\\color{red}{\\frac{du}{2}}}{\\color{red}{v}+1}<br \/>&amp; \\color{gray}{\\text{[Apply the substitution]}} \\\\[5pt]<br \/>&amp;= \\texttt{arctan}(2x+1) \\left[ x+\\frac{1}{2} \\right] \\:- \\frac{1}{4}\\int\\frac{du}{v+1}<br \/>&amp; \\color{gray}{\\text{[Rearrange the expression]}} \\\\[5pt]<br \/>&amp;= \\texttt{arctan}(2x+1) \\left[ x+\\frac{1}{2} \\right] \\:- \\frac{1}{4}\\texttt{ln}|v+1|\\:+ C<br \/>&amp; \\color{gray}{\\text{[Integrate]}} \\\\[5pt]<br \/>&amp;= \\texttt{arctan}(2x+1) \\left[ x+\\frac{1}{2} \\right] \\:- \\frac{1}{4}\\texttt{ln}(u^2+1)\\:+ C<br \/>&amp; \\color{gray}{\\text{[Substitute back for v (ln argument always positive)]}} \\\\[5pt]<br \/>&amp;= \\color{green}{\\texttt{arctan}(2x+1) \\left[ x+\\frac{1}{2} \\right] \\:- \\frac{1}{4}\\texttt{ln}((2x+1)^2+1)\\:+C}<br \/>&amp; \\color{gray}{\\text{[Substitute back for x]}} \\\\[5pt]<br \/>\\end{align*}<br \/>\\)<p>\u00a0<\/p><\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t\t\t\t\t<script type=\"application\/ld+json\">{\"@context\":\"https:\\\/\\\/schema.org\",\"@type\":\"FAQPage\",\"mainEntity\":[{\"@type\":\"Question\",\"name\":\"Problem 3: Solution\",\"acceptedAnswer\":{\"@type\":\"Answer\",\"text\":\"<p><span style=\\\"font-size: 18.6667px;\\\">Let \\\\(u=\\\\texttt{arctan}(2x+1)\\\\), so \\\\(du=\\\\frac{2}{(2x+1)^2+1}dx\\\\); and \\\\(dv=dx\\\\), so \\\\(v=x\\\\).<\\\/span><span style=\\\"font-size: 18.6667px;\\\">\\u00a0<\\\/span><span style=\\\"font-size: 18.6667px;\\\">Substituting these values into \\\\(uv\\\\:- \\\\int vdu\\\\) leads to the following:<\\\/span><\\\/p>\\\\(\\\\normalsize<br \\\/>\\\\begin{align*}\\\\displaystyle<br \\\/>\\\\int \\\\texttt{arctan}(2x+1)dx<br \\\/>&amp;= \\\\color{red}{\\\\underbrace{\\\\color{black}{\\\\texttt{arctan}(2x+1)}}_{\\\\large u}} \\\\color{red}{\\\\underbrace{\\\\color{black}{x}}_{\\\\large v}}\\\\:- \\\\int \\\\color{red}{\\\\underbrace{\\\\color{black}{x}}_{\\\\large v}} \\\\color{red}{\\\\underbrace{\\\\color{black}{\\\\left(\\\\frac{2}{(2x+1)^2+1}\\\\right)}}_{\\\\large du}}dx<br \\\/>&amp; \\\\color{gray}{\\\\text{[Apply the substitution]}} \\\\\\\\[5pt]<br \\\/>&amp;= x\\\\texttt{arctan}(2x+1)\\\\:- 2\\\\int \\\\frac{x}{(2x+1)^2+1}dx<br \\\/>&amp; \\\\color{gray}{\\\\text{[Rearrange the expression]}} \\\\\\\\[5pt]<br \\\/>\\\\end{align*}<br \\\/>\\\\)<p><span style=\\\"font-size: 18.6667px;\\\">\\u00a0<\\\/span><\\\/p><p><span style=\\\"font-size: 18.6667px;\\\">Use U-Sub with the given integrand. Let \\\\(u=2x+1\\\\), so \\\\(du=2dx\\\\); and \\\\(x=\\\\frac{u-1}{2}\\\\). While working the only integral left, it&#8217;s fine to have \\\\(u\\\\) and \\\\(x\\\\) at the same time given that the integrand is expressed in regards of just one variable. Substituting accordingly yields:<\\\/span><\\\/p>\\\\(\\\\normalsize<br \\\/>\\\\begin{align*}\\\\displaystyle<br \\\/>x\\\\texttt{arctan}(2x+1)\\\\:- 2\\\\int \\\\frac{x}{(2x+1)^2+1}dx<br \\\/>&amp;= x\\\\texttt{arctan}(2x+1)\\\\:- 2\\\\int \\\\left( \\\\frac{\\\\color{red}{\\\\left( \\\\frac{u-1}{2} \\\\right)}}{\\\\color{red}{u^2}+1} \\\\right) \\\\color{red}{\\\\left( \\\\frac{du}{2} \\\\right)}<br \\\/>&amp; \\\\color{gray}{\\\\text{[Apply the substitution]}} \\\\\\\\[5pt]<br \\\/>&amp;= x\\\\texttt{arctan}(2x+1)\\\\:- \\\\frac{1}{2}\\\\int\\\\frac{u-1}{u^2+1}du<br \\\/>&amp; \\\\color{gray}{\\\\text{[Rearrange the expression]}} \\\\\\\\[5pt]<br \\\/>&amp;= x\\\\texttt{arctan}(2x+1)\\\\:- \\\\frac{1}{2}\\\\left( \\\\int\\\\frac{u}{u^2+1}du\\\\:-\\\\int\\\\frac{1}{u^2+1}du \\\\right)<br \\\/>&amp; \\\\color{gray}{\\\\text{[Split the integrals strategically]}} \\\\\\\\[5pt]<br \\\/>&amp;= x\\\\texttt{arctan}(2x+1)\\\\:- \\\\frac{1}{2}\\\\int\\\\frac{u}{u^2+1}du\\\\:+\\\\frac{1}{2}\\\\int\\\\frac{1}{u^2+1}du<br \\\/>&amp; \\\\color{gray}{\\\\text{[Distribute the fraction]}} \\\\\\\\[5pt]<br \\\/>&amp;= x\\\\texttt{arctan}(2x+1)\\\\:+ \\\\frac{1}{2}\\\\texttt{arctan}(u)\\\\:- \\\\frac{1}{2}\\\\int\\\\frac{u}{u^2+1}du<br \\\/>&amp; \\\\color{gray}{\\\\text{[Integrate the second integral]}} \\\\\\\\[5pt]<br \\\/>&amp;= x\\\\texttt{arctan}(2x+1)\\\\:+ \\\\frac{1}{2}\\\\texttt{arctan}(2x+1)\\\\:- \\\\frac{1}{2}\\\\int\\\\frac{u}{u^2+1}du<br \\\/>&amp; \\\\color{gray}{\\\\text{[Substitute x back to the recently integrated term]}} \\\\\\\\[5pt]<br \\\/>&amp;= \\\\texttt{arctan}(2x+1) \\\\left[ x+\\\\frac{1}{2} \\\\right] \\\\:- \\\\frac{1}{2}\\\\int\\\\frac{u}{u^2+1}du<br \\\/>&amp; \\\\color{gray}{\\\\text{[Factor the arctan]}} \\\\\\\\[5pt]<br \\\/>\\\\end{align*}<br \\\/>\\\\)<p><span style=\\\"font-size: 18.6667px;\\\">\\u00a0<\\\/span><\\\/p><p><span style=\\\"font-size: 18.6667px;\\\">Use U-Sub with the given integrand. Let \\\\(v=u^2\\\\), so \\\\(dv=2udu\\\\)<\\\/span><span style=\\\"font-size: 18.6667px;\\\">. Substituting accordingly yields:<\\\/span><\\\/p>\\\\(\\\\normalsize<br \\\/>\\\\begin{align*}\\\\displaystyle<br \\\/>\\\\texttt{arctan}(2x+1) \\\\left[ x+\\\\frac{1}{2} \\\\right] \\\\:- \\\\frac{1}{2}\\\\int\\\\frac{u}{u^2+1}du<br \\\/>&amp;= \\\\texttt{arctan}(2x+1) \\\\left[ x+\\\\frac{1}{2} \\\\right] \\\\:- \\\\frac{1}{2}\\\\int\\\\frac{\\\\color{red}{\\\\frac{du}{2}}}{\\\\color{red}{v}+1}<br \\\/>&amp; \\\\color{gray}{\\\\text{[Apply the substitution]}} \\\\\\\\[5pt]<br \\\/>&amp;= \\\\texttt{arctan}(2x+1) \\\\left[ x+\\\\frac{1}{2} \\\\right] \\\\:- \\\\frac{1}{4}\\\\int\\\\frac{du}{v+1}<br \\\/>&amp; \\\\color{gray}{\\\\text{[Rearrange the expression]}} \\\\\\\\[5pt]<br \\\/>&amp;= \\\\texttt{arctan}(2x+1) \\\\left[ x+\\\\frac{1}{2} \\\\right] \\\\:- \\\\frac{1}{4}\\\\texttt{ln}|v+1|\\\\:+ C<br \\\/>&amp; \\\\color{gray}{\\\\text{[Integrate]}} \\\\\\\\[5pt]<br \\\/>&amp;= \\\\texttt{arctan}(2x+1) \\\\left[ x+\\\\frac{1}{2} \\\\right] \\\\:- \\\\frac{1}{4}\\\\texttt{ln}(u^2+1)\\\\:+ C<br \\\/>&amp; \\\\color{gray}{\\\\text{[Substitute back for v (ln argument always positive)]}} \\\\\\\\[5pt]<br \\\/>&amp;= \\\\color{green}{\\\\texttt{arctan}(2x+1) \\\\left[ x+\\\\frac{1}{2} \\\\right] \\\\:- \\\\frac{1}{4}\\\\texttt{ln}((2x+1)^2+1)\\\\:+C}<br \\\/>&amp; \\\\color{gray}{\\\\text{[Substitute back for x]}} \\\\\\\\[5pt]<br \\\/>\\\\end{align*}<br \\\/>\\\\)<p>\\u00a0<\\\/p>\"}}]}<\/script>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t<section class=\"elementor-section elementor-top-section elementor-element elementor-element-11ec88 elementor-section-boxed elementor-section-height-default elementor-section-height-default\" data-id=\"11ec88\" data-element_type=\"section\">\n\t\t\t\t\t\t<div class=\"elementor-container elementor-column-gap-default\">\n\t\t\t\t\t<div class=\"elementor-column elementor-col-100 elementor-top-column elementor-element elementor-element-20af7f46\" data-id=\"20af7f46\" data-element_type=\"column\">\n\t\t\t<div class=\"elementor-widget-wrap elementor-element-populated\">\n\t\t\t\t\t\t<div class=\"elementor-element elementor-element-569b9ca1 elementor-widget elementor-widget-toggle\" data-id=\"569b9ca1\" data-element_type=\"widget\" data-widget_type=\"toggle.default\">\n\t\t\t\t<div class=\"elementor-widget-container\">\n\t\t\t\t\t\t\t<div class=\"elementor-toggle\">\n\t\t\t\t\t\t\t<div class=\"elementor-toggle-item\">\n\t\t\t\t\t<h2 id=\"elementor-tab-title-1451\" class=\"elementor-tab-title\" data-tab=\"1\" role=\"button\" aria-controls=\"elementor-tab-content-1451\" aria-expanded=\"false\">\n\t\t\t\t\t\t\t\t\t\t\t\t<span class=\"elementor-toggle-icon elementor-toggle-icon-left\" aria-hidden=\"true\">\n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t<span class=\"elementor-toggle-icon-closed\"><i class=\"fas fa-toggle-off\"><\/i><\/span>\n\t\t\t\t\t\t\t\t<span class=\"elementor-toggle-icon-opened\"><i class=\"elementor-toggle-icon-opened fas fa-toggle-on\"><\/i><\/span>\n\t\t\t\t\t\t\t\t\t\t\t\t\t<\/span>\n\t\t\t\t\t\t\t\t\t\t\t\t<a class=\"elementor-toggle-title\" tabindex=\"0\">Problem 4: Solution<\/a>\n\t\t\t\t\t<\/h2>\n\n\t\t\t\t\t<div id=\"elementor-tab-content-1451\" class=\"elementor-tab-content elementor-clearfix\" data-tab=\"1\" role=\"region\" aria-labelledby=\"elementor-tab-title-1451\"><p><span style=\"font-size: 18.6667px;\">Let \\(u=\\texttt{cos}(\\sqrt{x})\\), so \\(du=\\frac{-\\texttt{sin}(\\sqrt{x})}{2\\sqrt{x}}dx\\); and \\(dv=dx\\), so \\(v=x\\).<\/span><span style=\"font-size: 18.6667px;\">\u00a0<\/span><span style=\"font-size: 18.6667px;\">Substituting these values into \\(uv &#8211; \\int vdu\\) leads to the following:<\/span><\/p>\\(\\normalsize<br \/>\\begin{align*}\\displaystyle<br \/>\\int \\texttt{cos}{\\sqrt{x}}dx<br \/>&amp;= \\color{red}{\\underbrace{\\color{black}{\\texttt{cos}(\\sqrt{x})}}_{\\large u}} \\color{red}{\\underbrace{\\color{black}{x}}_{\\large v}}\\:- \\int\\ \\color{red}{\\underbrace{\\color{black}{x}}_{\\large v}} \\color{red}{\\underbrace{\\color{black}{\\left(\\frac{-\\texttt{sin}(\\sqrt{x})}{2\\sqrt{x}}\\right)}}_{\\large du}}dx<br \/>&amp; \\color{gray}{\\text{[Apply the substitution]}} \\\\[5pt]<br \/>&amp;= x\\texttt{cos}(\\sqrt{x})\\:+ \\frac{1}{2}\\int\\frac{x\\texttt{sin}(\\sqrt{x})}{\\sqrt{x}}dx<br \/>&amp; \\color{gray}{\\text{[Rearrange the expression]}} \\\\[5pt]<br \/>&amp;= x\\texttt{cos}(\\sqrt{x})\\:+ \\frac{1}{2}\\int\\sqrt{x}\\texttt{sin}(\\sqrt{x})dx<br \/>&amp; \\color{gray}{\\text{[Simplify terms]}} \\\\[5pt]<br \/>\\end{align*}<br \/>\\)<p>\u00a0<\/p><p>\u00a0<\/p><p><span style=\"font-size: 18.6667px;\">Use U-Sub with the given integrand. Let \\(u=\\sqrt{x}\\), so \\(du=\\frac{1}{2\\sqrt{x}}dx\\); \\(dx=2\\sqrt{x}du=2udu\\)<\/span><span style=\"font-size: 18.6667px;\">. Substituting accordingly yields:<\/span><\/p>\\(\\normalsize<br \/>\\begin{align*}\\displaystyle<br \/>x\\texttt{cos}(\\sqrt{x})\\:+ \\frac{1}{2}\\int\\sqrt{x}\\texttt{sin}(\\sqrt{x})dx<br \/>&amp;= x\\texttt{cos}(\\sqrt{x})\\:+ \\frac{1}{2}\\int \\color{red}{\\underbrace{\\color{black}{u}}_{\\large u^2}} \\texttt{sin}(\\color{red}{\\underbrace{\\color{black}{u}}_{\\large u}})\\color{red}{(2udu)}<br \/>&amp; \\color{gray}{\\text{[Apply the substitution]}} \\\\[5pt]<br \/>&amp;= x\\texttt{cos}(\\sqrt{x})\\:+ \\int u^2\\texttt{sin}(u)du<br \/>&amp; \\color{gray}{\\text{[Rearrange the expression]}} \\\\[5pt]<br \/>\\end{align*}<br \/>\\)<p>\u00a0<\/p><p>\u00a0<\/p><p><span style=\"font-size: 18.6667px;\">Apply IBP again. Let \\(m=u^2\\), so \\(dm=2udu\\); and \\(dw=\\texttt{sin}(u)du\\), so \\(w=-\\texttt{cos}(u)\\).<\/span><span style=\"font-size: 18.6667px;\">\u00a0<\/span><span style=\"font-size: 18.6667px;\">Substituting these values into \\(mw\\:- \\int wdm\\) leads to the following:<\/span><span style=\"font-size: 18.6667px;\"><br \/><\/span><\/p>\\(\\normalsize<br \/>\\begin{align*}\\displaystyle<br \/>x\\texttt{cos}(\\sqrt{x})\\:+ \\int u^2\\texttt{sin}(u)du<br \/>&amp;= x\\texttt{cos}(\\sqrt{x})\\:+ \\left[ \\left( \\color{red}{\\underbrace{\\color{black}{u^2}}_{\\large m}}\\right) \\left( \\color{red}{\\underbrace{\\color{black}{-\\texttt{cos}(u)}}_{\\large w}} \\right)\\: -\\int \\left( \\color{red}{\\underbrace{\\color{black}{-\\texttt{cos}(u)}}_{\\large w}}\\right)\u00a0 \\left( \\color{red}{\\underbrace{\\color{black}{2udu}}_{\\large dm}} \\right)\u00a0 \\right]<br \/>&amp; \\color{gray}{\\text{[Apply the substitution]}} \\\\[5pt]<br \/>&amp;= x\\texttt{cos}(\\sqrt{x})\\:- u^2\\texttt{cos}(u)\\:+ 2\\int u\\texttt{cos}(u)du<br \/>&amp; \\color{gray}{\\text{[Rearrange the expression]}} \\\\[5pt]<br \/>\\end{align*}<br \/>\\)<p>\u00a0<\/p><p>\u00a0<\/p><p><span style=\"font-size: 18.6667px;\">Apply IBP again. Let \\(m=u\\), so \\(dm=du\\); and \\(dw=\\texttt{cos}(u)du\\), so \\(w=\\texttt{sin}(u)\\).<\/span><span style=\"font-size: 18.6667px;\">\u00a0<\/span><span style=\"font-size: 18.6667px;\">Substituting these values into \\(mw\\:- \\int wdm\\) leads to the following:<\/span><\/p>\\(\\normalsize<br \/>\\begin{align*}\\displaystyle<br \/>x\\texttt{cos}(\\sqrt{x})\\: &amp;- u^2\\texttt{cos}(u)\\; + 2\\int u\\texttt{cos}(u)du\\\\[5pt]<br \/>&amp;= \\texttt{cos}(\\sqrt{x})\\:- u^2\\texttt{cos}(u)\\:+ 2\\left( (\\color{red}{\\underbrace{\\color{black}{u}}_{\\large m}})(\\color{red}{\\underbrace{\\color{black}{\\texttt{sin}(u)}}_{\\large w}})\\:- \\int \\color{red}{\\underbrace{\\color{black}{\\texttt{sin}(u)}}_{\\large w}} \\color{red}{\\underbrace{\\color{black}{du}}_{dm}} \\right)<br \/>&amp; \\color{gray}{\\text{[Apply the substitution]}} \\\\[5pt]<br \/>&amp;= x\\texttt{cos}(\\sqrt{x})\\:- u^2\\texttt{cos}(u)\\:+ 2u\\texttt{sin}(u)\\:- 2\\int \\texttt{sin}(u)du<br \/>&amp; \\color{gray}{\\text{[Rearrange the expression]}} \\\\[5pt]<br \/>&amp;= x\\texttt{cos}(\\sqrt{x})\\:- u^2\\texttt{cos}(u)\\:+ 2u\\texttt{sin}(u)\\:+ 2\\texttt{cos}(u)\\:+ C<br \/>&amp; \\color{gray}{\\text{[Integrate]}} \\\\[5pt]<br \/>&amp;= x\\texttt{cos}(\\sqrt{x})\\:- x\\texttt{cos}(\\sqrt{x})\\:+ 2\\sqrt{x}\\texttt{sin}(\\sqrt{x})\\:+ 2\\texttt{cos}(\\sqrt{x})\\:+ C<br \/>&amp; \\color{gray}{\\text{[Substitute back to x]}} \\\\[5pt]<br \/>&amp;= \\color{green}{2\\sqrt{x}\\texttt{sin}(\\sqrt{x})\\:+ 2\\texttt{cos}(\\sqrt{x})\\:+ C}<br \/>&amp; \\color{gray}{\\text{[Simplify]}} \\\\[5pt]<br \/>\\end{align*}<br \/>\\)<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t\t\t\t\t<script type=\"application\/ld+json\">{\"@context\":\"https:\\\/\\\/schema.org\",\"@type\":\"FAQPage\",\"mainEntity\":[{\"@type\":\"Question\",\"name\":\"Problem 4: Solution\",\"acceptedAnswer\":{\"@type\":\"Answer\",\"text\":\"<p><span style=\\\"font-size: 18.6667px;\\\">Let \\\\(u=\\\\texttt{cos}(\\\\sqrt{x})\\\\), so \\\\(du=\\\\frac{-\\\\texttt{sin}(\\\\sqrt{x})}{2\\\\sqrt{x}}dx\\\\); and \\\\(dv=dx\\\\), so \\\\(v=x\\\\).<\\\/span><span style=\\\"font-size: 18.6667px;\\\">\\u00a0<\\\/span><span style=\\\"font-size: 18.6667px;\\\">Substituting these values into \\\\(uv &#8211; \\\\int vdu\\\\) leads to the following:<\\\/span><\\\/p>\\\\(\\\\normalsize<br \\\/>\\\\begin{align*}\\\\displaystyle<br \\\/>\\\\int \\\\texttt{cos}{\\\\sqrt{x}}dx<br \\\/>&amp;= \\\\color{red}{\\\\underbrace{\\\\color{black}{\\\\texttt{cos}(\\\\sqrt{x})}}_{\\\\large u}} \\\\color{red}{\\\\underbrace{\\\\color{black}{x}}_{\\\\large v}}\\\\:- \\\\int\\\\ \\\\color{red}{\\\\underbrace{\\\\color{black}{x}}_{\\\\large v}} \\\\color{red}{\\\\underbrace{\\\\color{black}{\\\\left(\\\\frac{-\\\\texttt{sin}(\\\\sqrt{x})}{2\\\\sqrt{x}}\\\\right)}}_{\\\\large du}}dx<br \\\/>&amp; \\\\color{gray}{\\\\text{[Apply the substitution]}} \\\\\\\\[5pt]<br \\\/>&amp;= x\\\\texttt{cos}(\\\\sqrt{x})\\\\:+ \\\\frac{1}{2}\\\\int\\\\frac{x\\\\texttt{sin}(\\\\sqrt{x})}{\\\\sqrt{x}}dx<br \\\/>&amp; \\\\color{gray}{\\\\text{[Rearrange the expression]}} \\\\\\\\[5pt]<br \\\/>&amp;= x\\\\texttt{cos}(\\\\sqrt{x})\\\\:+ \\\\frac{1}{2}\\\\int\\\\sqrt{x}\\\\texttt{sin}(\\\\sqrt{x})dx<br \\\/>&amp; \\\\color{gray}{\\\\text{[Simplify terms]}} \\\\\\\\[5pt]<br \\\/>\\\\end{align*}<br \\\/>\\\\)<p>\\u00a0<\\\/p><p>\\u00a0<\\\/p><p><span style=\\\"font-size: 18.6667px;\\\">Use U-Sub with the given integrand. Let \\\\(u=\\\\sqrt{x}\\\\), so \\\\(du=\\\\frac{1}{2\\\\sqrt{x}}dx\\\\); \\\\(dx=2\\\\sqrt{x}du=2udu\\\\)<\\\/span><span style=\\\"font-size: 18.6667px;\\\">. Substituting accordingly yields:<\\\/span><\\\/p>\\\\(\\\\normalsize<br \\\/>\\\\begin{align*}\\\\displaystyle<br \\\/>x\\\\texttt{cos}(\\\\sqrt{x})\\\\:+ \\\\frac{1}{2}\\\\int\\\\sqrt{x}\\\\texttt{sin}(\\\\sqrt{x})dx<br \\\/>&amp;= x\\\\texttt{cos}(\\\\sqrt{x})\\\\:+ \\\\frac{1}{2}\\\\int \\\\color{red}{\\\\underbrace{\\\\color{black}{u}}_{\\\\large u^2}} \\\\texttt{sin}(\\\\color{red}{\\\\underbrace{\\\\color{black}{u}}_{\\\\large u}})\\\\color{red}{(2udu)}<br \\\/>&amp; \\\\color{gray}{\\\\text{[Apply the substitution]}} \\\\\\\\[5pt]<br \\\/>&amp;= x\\\\texttt{cos}(\\\\sqrt{x})\\\\:+ \\\\int u^2\\\\texttt{sin}(u)du<br \\\/>&amp; \\\\color{gray}{\\\\text{[Rearrange the expression]}} \\\\\\\\[5pt]<br \\\/>\\\\end{align*}<br \\\/>\\\\)<p>\\u00a0<\\\/p><p>\\u00a0<\\\/p><p><span style=\\\"font-size: 18.6667px;\\\">Apply IBP again. Let \\\\(m=u^2\\\\), so \\\\(dm=2udu\\\\); and \\\\(dw=\\\\texttt{sin}(u)du\\\\), so \\\\(w=-\\\\texttt{cos}(u)\\\\).<\\\/span><span style=\\\"font-size: 18.6667px;\\\">\\u00a0<\\\/span><span style=\\\"font-size: 18.6667px;\\\">Substituting these values into \\\\(mw\\\\:- \\\\int wdm\\\\) leads to the following:<\\\/span><span style=\\\"font-size: 18.6667px;\\\"><br \\\/><\\\/span><\\\/p>\\\\(\\\\normalsize<br \\\/>\\\\begin{align*}\\\\displaystyle<br \\\/>x\\\\texttt{cos}(\\\\sqrt{x})\\\\:+ \\\\int u^2\\\\texttt{sin}(u)du<br \\\/>&amp;= x\\\\texttt{cos}(\\\\sqrt{x})\\\\:+ \\\\left[ \\\\left( \\\\color{red}{\\\\underbrace{\\\\color{black}{u^2}}_{\\\\large m}}\\\\right) \\\\left( \\\\color{red}{\\\\underbrace{\\\\color{black}{-\\\\texttt{cos}(u)}}_{\\\\large w}} \\\\right)\\\\: -\\\\int \\\\left( \\\\color{red}{\\\\underbrace{\\\\color{black}{-\\\\texttt{cos}(u)}}_{\\\\large w}}\\\\right)\\u00a0 \\\\left( \\\\color{red}{\\\\underbrace{\\\\color{black}{2udu}}_{\\\\large dm}} \\\\right)\\u00a0 \\\\right]<br \\\/>&amp; \\\\color{gray}{\\\\text{[Apply the substitution]}} \\\\\\\\[5pt]<br \\\/>&amp;= x\\\\texttt{cos}(\\\\sqrt{x})\\\\:- u^2\\\\texttt{cos}(u)\\\\:+ 2\\\\int u\\\\texttt{cos}(u)du<br \\\/>&amp; \\\\color{gray}{\\\\text{[Rearrange the expression]}} \\\\\\\\[5pt]<br \\\/>\\\\end{align*}<br \\\/>\\\\)<p>\\u00a0<\\\/p><p>\\u00a0<\\\/p><p><span style=\\\"font-size: 18.6667px;\\\">Apply IBP again. Let \\\\(m=u\\\\), so \\\\(dm=du\\\\); and \\\\(dw=\\\\texttt{cos}(u)du\\\\), so \\\\(w=\\\\texttt{sin}(u)\\\\).<\\\/span><span style=\\\"font-size: 18.6667px;\\\">\\u00a0<\\\/span><span style=\\\"font-size: 18.6667px;\\\">Substituting these values into \\\\(mw\\\\:- \\\\int wdm\\\\) leads to the following:<\\\/span><\\\/p>\\\\(\\\\normalsize<br \\\/>\\\\begin{align*}\\\\displaystyle<br \\\/>x\\\\texttt{cos}(\\\\sqrt{x})\\\\: &amp;- u^2\\\\texttt{cos}(u)\\\\; + 2\\\\int u\\\\texttt{cos}(u)du\\\\\\\\[5pt]<br \\\/>&amp;= \\\\texttt{cos}(\\\\sqrt{x})\\\\:- u^2\\\\texttt{cos}(u)\\\\:+ 2\\\\left( (\\\\color{red}{\\\\underbrace{\\\\color{black}{u}}_{\\\\large m}})(\\\\color{red}{\\\\underbrace{\\\\color{black}{\\\\texttt{sin}(u)}}_{\\\\large w}})\\\\:- \\\\int \\\\color{red}{\\\\underbrace{\\\\color{black}{\\\\texttt{sin}(u)}}_{\\\\large w}} \\\\color{red}{\\\\underbrace{\\\\color{black}{du}}_{dm}} \\\\right)<br \\\/>&amp; \\\\color{gray}{\\\\text{[Apply the substitution]}} \\\\\\\\[5pt]<br \\\/>&amp;= x\\\\texttt{cos}(\\\\sqrt{x})\\\\:- u^2\\\\texttt{cos}(u)\\\\:+ 2u\\\\texttt{sin}(u)\\\\:- 2\\\\int \\\\texttt{sin}(u)du<br \\\/>&amp; \\\\color{gray}{\\\\text{[Rearrange the expression]}} \\\\\\\\[5pt]<br \\\/>&amp;= x\\\\texttt{cos}(\\\\sqrt{x})\\\\:- u^2\\\\texttt{cos}(u)\\\\:+ 2u\\\\texttt{sin}(u)\\\\:+ 2\\\\texttt{cos}(u)\\\\:+ C<br \\\/>&amp; \\\\color{gray}{\\\\text{[Integrate]}} \\\\\\\\[5pt]<br \\\/>&amp;= x\\\\texttt{cos}(\\\\sqrt{x})\\\\:- x\\\\texttt{cos}(\\\\sqrt{x})\\\\:+ 2\\\\sqrt{x}\\\\texttt{sin}(\\\\sqrt{x})\\\\:+ 2\\\\texttt{cos}(\\\\sqrt{x})\\\\:+ C<br \\\/>&amp; \\\\color{gray}{\\\\text{[Substitute back to x]}} \\\\\\\\[5pt]<br \\\/>&amp;= \\\\color{green}{2\\\\sqrt{x}\\\\texttt{sin}(\\\\sqrt{x})\\\\:+ 2\\\\texttt{cos}(\\\\sqrt{x})\\\\:+ C}<br \\\/>&amp; \\\\color{gray}{\\\\text{[Simplify]}} \\\\\\\\[5pt]<br \\\/>\\\\end{align*}<br \\\/>\\\\)\"}}]}<\/script>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t<section class=\"elementor-section elementor-top-section elementor-element elementor-element-36e16591 elementor-section-boxed elementor-section-height-default elementor-section-height-default\" data-id=\"36e16591\" data-element_type=\"section\">\n\t\t\t\t\t\t<div class=\"elementor-container elementor-column-gap-default\">\n\t\t\t\t\t<div class=\"elementor-column elementor-col-100 elementor-top-column elementor-element elementor-element-3492158f\" data-id=\"3492158f\" data-element_type=\"column\">\n\t\t\t<div class=\"elementor-widget-wrap elementor-element-populated\">\n\t\t\t\t\t\t<div class=\"elementor-element elementor-element-4da605f1 elementor-widget elementor-widget-toggle\" data-id=\"4da605f1\" data-element_type=\"widget\" data-widget_type=\"toggle.default\">\n\t\t\t\t<div class=\"elementor-widget-container\">\n\t\t\t\t\t\t\t<div class=\"elementor-toggle\">\n\t\t\t\t\t\t\t<div class=\"elementor-toggle-item\">\n\t\t\t\t\t<h2 id=\"elementor-tab-title-1301\" class=\"elementor-tab-title\" data-tab=\"1\" role=\"button\" aria-controls=\"elementor-tab-content-1301\" aria-expanded=\"false\">\n\t\t\t\t\t\t\t\t\t\t\t\t<span class=\"elementor-toggle-icon elementor-toggle-icon-left\" aria-hidden=\"true\">\n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t<span class=\"elementor-toggle-icon-closed\"><i class=\"fas fa-toggle-off\"><\/i><\/span>\n\t\t\t\t\t\t\t\t<span class=\"elementor-toggle-icon-opened\"><i class=\"elementor-toggle-icon-opened fas fa-toggle-on\"><\/i><\/span>\n\t\t\t\t\t\t\t\t\t\t\t\t\t<\/span>\n\t\t\t\t\t\t\t\t\t\t\t\t<a class=\"elementor-toggle-title\" tabindex=\"0\">Problem 5: Solution<\/a>\n\t\t\t\t\t<\/h2>\n\n\t\t\t\t\t<div id=\"elementor-tab-content-1301\" class=\"elementor-tab-content elementor-clearfix\" data-tab=\"1\" role=\"region\" aria-labelledby=\"elementor-tab-title-1301\"><p><span style=\"font-size: 18.6667px;\">Let \\(u=\\texttt{e}^{\\sqrt{x}}\\), so \\(du=\\frac{\\texttt{e}^{\\sqrt{x}}}{2\\sqrt{x}}dx\\); and \\(dv=dx\\), so \\(v=x\\).<\/span><span style=\"font-size: 18.6667px;\">\u00a0<\/span><span style=\"font-size: 18.6667px;\">Substituting these values into \\(uv\\:- \\int vdu\\) leads to the following:<\/span><\/p>\\(\\normalsize<br \/>\\begin{align*}\\displaystyle<br \/>\\int \\texttt{e}^{\\sqrt{x}}dx<br \/>&amp;= \\color{red}{\\underbrace{\\color{black}{\\left( \\texttt{e}^{\\sqrt{x}} \\right)}}_{\\large u}} \\color{red}{\\underbrace{\\color{black}{x}}_{\\large v}}\\:- \\int \\color{red}{\\underbrace{\\color{black}{x}}_{\\large v}} \\color{red}{\\underbrace{\\color{black}{\\left(\u00a0\\frac{\\texttt{e}^{\\sqrt{x}}}{2\\sqrt{x}}dx \\right)}}_{\\large du}}<br \/>&amp; \\color{gray}{\\text{[Apply the substitution]}} \\\\[5pt]<br \/>&amp;= x\\texttt{e}^{\\sqrt{x}}\\:- \\frac{1}{2}\\int\\frac{x\\texttt{e}^{\\sqrt{x}}}{\\sqrt{x}}dx<br \/>&amp; \\color{gray}{\\text{[Rearrange the expression]}} \\\\[5pt]<br \/>&amp;= x\\texttt{e}^{\\sqrt{x}}\\:- \\frac{1}{2}\\int \\sqrt{x}\\texttt{e}^{\\sqrt{x}}dx<br \/>&amp; \\color{gray}{\\text{[Simplify]}} \\\\[5pt]<br \/>\\end{align*}<br \/>\\)<p>\u00a0<\/p><p>\u00a0<\/p><p><span style=\"font-size: 18.6667px;\">Use U-Sub with the given integrand. Let \\(u=\\sqrt{x}\\), so \\(du=\\frac{1}{2\\sqrt{x}}dx\\); \\(dx=2\\sqrt{x}du=2udu\\)<\/span><span style=\"font-size: 18.6667px;\">. Substituting accordingly yields:<\/span><\/p>\\(\\normalsize<br \/>\\begin{align*}\\displaystyle<br \/>x\\texttt{e}^{\\sqrt{x}}\\:- \\frac{1}{2}\\int \\sqrt{x}\\texttt{e}^{\\sqrt{x}}dx<br \/>&amp;= x\\texttt{e}^{\\sqrt{x}}\\:- \\frac{1}{2}\\int \\color{red}{2u^2\\texttt{e}^udu}<br \/>&amp; \\color{gray}{\\text{[Apply the substitution]}} \\\\[5pt]<br \/>&amp;= x\\texttt{e}^{\\sqrt{x}}\\:- \\int u^2\\texttt{e}^udu<br \/>&amp; \\color{gray}{\\text{[Simplify]}} \\\\[5pt]<br \/>\\end{align*}<br \/>\\)<p>\u00a0<\/p><p>\u00a0<\/p><p><span style=\"font-size: 18.6667px;\">Apply IBP. Let \\(m=u^2\\), so \\(dm=2udu\\); and \\(dw=\\texttt{e}^udu\\), so \\(w=\\texttt{e}^u\\).<\/span><span style=\"font-size: 18.6667px;\">\u00a0<\/span><span style=\"font-size: 18.6667px;\">Substituting these values into \\(mw\\:- \\int wdm\\) leads to the following:<br \/><\/span><\/p>\\(\\normalsize<br \/>\\begin{align*}\\displaystyle<br \/>x\\texttt{e}^{\\sqrt{x}}\\:- \\int u^2\\texttt{e}^udu<br \/>&amp;= x\\texttt{e}^{\\sqrt{x}}\\:- \\left( \\color{red}{\\underbrace{\\color{black}{u^2}}_{\\large m}} \\color{red}{\\underbrace{\\color{black}{\\texttt{e}^u}}_{\\large w}}\\:- \\int \\color{red}{\\underbrace{\\color{black}{\\texttt{e}^u}}_{\\large w}} \\color{red}{\\underbrace{\\color{black}{(2udu)}}_{dm}} \\right)<br \/>&amp; \\color{gray}{\\text{[Apply the substitution]}} \\\\[5pt]<br \/>&amp;= x\\texttt{e}^{\\sqrt{x}}\\:- u^2\\texttt{e}^u\\:+ 2\\int u\\texttt{e}^udu<br \/>&amp; \\color{gray}{\\text{[Rearrange the expression]}} \\\\[5pt]<br \/>\\end{align*}<br \/>\\)<p>\u00a0<\/p><p>\u00a0<\/p><p><span style=\"font-size: 18.6667px;\">Apply IBP one last time. Let \\(m=u\\), so \\(dm=du\\); and \\(dw=e^udu\\), so \\(w=e^u\\).<\/span><span style=\"font-size: 18.6667px;\">\u00a0<\/span><span style=\"font-size: 18.6667px;\">Substituting these values into \\(mw\\:- \\int wdm\\) leads to the following:<\/span><\/p>\\(\\normalsize<br \/>\\begin{align*}\\displaystyle<br \/>x\\texttt{e}^{\\sqrt{x}}\\:- u^2\\texttt{e}^u\\:+ 2\\int u\\texttt{e}^udu<br \/>&amp;= x\\texttt{e}^{\\sqrt{x}}\\:- u^2\\texttt{e}^u\\:+ 2\\left( \\color{red}{\\underbrace{\\color{black}{u}}_{\\large m}} \\color{red}{\\underbrace{\\color{black}{\\texttt{e}^u}}_{\\large w}}\\:- \\int \\color{red}{\\underbrace{\\color{black}{\\texttt{e}^u}}_{\\large w}} \\color{red}{\\underbrace{\\color{black}{du}}_{dm}} \\right)<br \/>&amp; \\color{gray}{\\text{[Apply the substitution]}} \\\\[5pt]<br \/>&amp;= x\\texttt{e}^{\\sqrt{x}}\\:- u^2\\texttt{e}^u\\:+ 2u\\texttt{e}^u\\:- 2\\int \\texttt{e}^udu<br \/>&amp; \\color{gray}{\\text{[Rearrange the expression]}} \\\\[5pt]<br \/>&amp;= x\\texttt{e}^{\\sqrt{x}}\\:- u^2\\texttt{e}^u\\:+ 2u\\texttt{e}^u\\:- 2\\texttt{e}^u\\:+ C<br \/>&amp; \\color{gray}{\\text{[Intergate]}} \\\\[5pt]<br \/>&amp;= x\\texttt{e}^{\\sqrt{x}}\\:- x\\texttt{e}^{\\sqrt{x}}\\:+ 2\\sqrt{x}\\texttt{e}^{\\sqrt{x}}\\:- 2\\texttt{e}^{\\sqrt{x}}\\:+ C<br \/>&amp; \\color{gray}{\\text{[Substitute back to x]}} \\\\[5pt]<br \/>&amp;= 2\\sqrt{x}\\texttt{e}^{\\sqrt{x}}\\:- 2\\texttt{e}^{\\sqrt{x}}\\:+ C<br \/>&amp; \\color{gray}{\\text{[Simplify]}} \\\\[5pt]<br \/>&amp;= \\color{green}{2\\texttt{e}^{\\sqrt{x}}(\\sqrt{x}-1)\\:+ C}<br \/>&amp; \\color{gray}{\\text{[Factor]}} \\\\[5pt]<br \/>\\end{align*}<br \/>\\)<p>\u00a0<\/p><\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t\t\t\t\t<script type=\"application\/ld+json\">{\"@context\":\"https:\\\/\\\/schema.org\",\"@type\":\"FAQPage\",\"mainEntity\":[{\"@type\":\"Question\",\"name\":\"Problem 5: Solution\",\"acceptedAnswer\":{\"@type\":\"Answer\",\"text\":\"<p><span style=\\\"font-size: 18.6667px;\\\">Let \\\\(u=\\\\texttt{e}^{\\\\sqrt{x}}\\\\), so \\\\(du=\\\\frac{\\\\texttt{e}^{\\\\sqrt{x}}}{2\\\\sqrt{x}}dx\\\\); and \\\\(dv=dx\\\\), so \\\\(v=x\\\\).<\\\/span><span style=\\\"font-size: 18.6667px;\\\">\\u00a0<\\\/span><span style=\\\"font-size: 18.6667px;\\\">Substituting these values into \\\\(uv\\\\:- \\\\int vdu\\\\) leads to the following:<\\\/span><\\\/p>\\\\(\\\\normalsize<br \\\/>\\\\begin{align*}\\\\displaystyle<br \\\/>\\\\int \\\\texttt{e}^{\\\\sqrt{x}}dx<br \\\/>&amp;= \\\\color{red}{\\\\underbrace{\\\\color{black}{\\\\left( \\\\texttt{e}^{\\\\sqrt{x}} \\\\right)}}_{\\\\large u}} \\\\color{red}{\\\\underbrace{\\\\color{black}{x}}_{\\\\large v}}\\\\:- \\\\int \\\\color{red}{\\\\underbrace{\\\\color{black}{x}}_{\\\\large v}} \\\\color{red}{\\\\underbrace{\\\\color{black}{\\\\left(\\u00a0\\\\frac{\\\\texttt{e}^{\\\\sqrt{x}}}{2\\\\sqrt{x}}dx \\\\right)}}_{\\\\large du}}<br \\\/>&amp; \\\\color{gray}{\\\\text{[Apply the substitution]}} \\\\\\\\[5pt]<br \\\/>&amp;= x\\\\texttt{e}^{\\\\sqrt{x}}\\\\:- \\\\frac{1}{2}\\\\int\\\\frac{x\\\\texttt{e}^{\\\\sqrt{x}}}{\\\\sqrt{x}}dx<br \\\/>&amp; \\\\color{gray}{\\\\text{[Rearrange the expression]}} \\\\\\\\[5pt]<br \\\/>&amp;= x\\\\texttt{e}^{\\\\sqrt{x}}\\\\:- \\\\frac{1}{2}\\\\int \\\\sqrt{x}\\\\texttt{e}^{\\\\sqrt{x}}dx<br \\\/>&amp; \\\\color{gray}{\\\\text{[Simplify]}} \\\\\\\\[5pt]<br \\\/>\\\\end{align*}<br \\\/>\\\\)<p>\\u00a0<\\\/p><p>\\u00a0<\\\/p><p><span style=\\\"font-size: 18.6667px;\\\">Use U-Sub with the given integrand. Let \\\\(u=\\\\sqrt{x}\\\\), so \\\\(du=\\\\frac{1}{2\\\\sqrt{x}}dx\\\\); \\\\(dx=2\\\\sqrt{x}du=2udu\\\\)<\\\/span><span style=\\\"font-size: 18.6667px;\\\">. Substituting accordingly yields:<\\\/span><\\\/p>\\\\(\\\\normalsize<br \\\/>\\\\begin{align*}\\\\displaystyle<br \\\/>x\\\\texttt{e}^{\\\\sqrt{x}}\\\\:- \\\\frac{1}{2}\\\\int \\\\sqrt{x}\\\\texttt{e}^{\\\\sqrt{x}}dx<br \\\/>&amp;= x\\\\texttt{e}^{\\\\sqrt{x}}\\\\:- \\\\frac{1}{2}\\\\int \\\\color{red}{2u^2\\\\texttt{e}^udu}<br \\\/>&amp; \\\\color{gray}{\\\\text{[Apply the substitution]}} \\\\\\\\[5pt]<br \\\/>&amp;= x\\\\texttt{e}^{\\\\sqrt{x}}\\\\:- \\\\int u^2\\\\texttt{e}^udu<br \\\/>&amp; \\\\color{gray}{\\\\text{[Simplify]}} \\\\\\\\[5pt]<br \\\/>\\\\end{align*}<br \\\/>\\\\)<p>\\u00a0<\\\/p><p>\\u00a0<\\\/p><p><span style=\\\"font-size: 18.6667px;\\\">Apply IBP. Let \\\\(m=u^2\\\\), so \\\\(dm=2udu\\\\); and \\\\(dw=\\\\texttt{e}^udu\\\\), so \\\\(w=\\\\texttt{e}^u\\\\).<\\\/span><span style=\\\"font-size: 18.6667px;\\\">\\u00a0<\\\/span><span style=\\\"font-size: 18.6667px;\\\">Substituting these values into \\\\(mw\\\\:- \\\\int wdm\\\\) leads to the following:<br \\\/><\\\/span><\\\/p>\\\\(\\\\normalsize<br \\\/>\\\\begin{align*}\\\\displaystyle<br \\\/>x\\\\texttt{e}^{\\\\sqrt{x}}\\\\:- \\\\int u^2\\\\texttt{e}^udu<br \\\/>&amp;= x\\\\texttt{e}^{\\\\sqrt{x}}\\\\:- \\\\left( \\\\color{red}{\\\\underbrace{\\\\color{black}{u^2}}_{\\\\large m}} \\\\color{red}{\\\\underbrace{\\\\color{black}{\\\\texttt{e}^u}}_{\\\\large w}}\\\\:- \\\\int \\\\color{red}{\\\\underbrace{\\\\color{black}{\\\\texttt{e}^u}}_{\\\\large w}} \\\\color{red}{\\\\underbrace{\\\\color{black}{(2udu)}}_{dm}} \\\\right)<br \\\/>&amp; \\\\color{gray}{\\\\text{[Apply the substitution]}} \\\\\\\\[5pt]<br \\\/>&amp;= x\\\\texttt{e}^{\\\\sqrt{x}}\\\\:- u^2\\\\texttt{e}^u\\\\:+ 2\\\\int u\\\\texttt{e}^udu<br \\\/>&amp; \\\\color{gray}{\\\\text{[Rearrange the expression]}} \\\\\\\\[5pt]<br \\\/>\\\\end{align*}<br \\\/>\\\\)<p>\\u00a0<\\\/p><p>\\u00a0<\\\/p><p><span style=\\\"font-size: 18.6667px;\\\">Apply IBP one last time. Let \\\\(m=u\\\\), so \\\\(dm=du\\\\); and \\\\(dw=e^udu\\\\), so \\\\(w=e^u\\\\).<\\\/span><span style=\\\"font-size: 18.6667px;\\\">\\u00a0<\\\/span><span style=\\\"font-size: 18.6667px;\\\">Substituting these values into \\\\(mw\\\\:- \\\\int wdm\\\\) leads to the following:<\\\/span><\\\/p>\\\\(\\\\normalsize<br \\\/>\\\\begin{align*}\\\\displaystyle<br \\\/>x\\\\texttt{e}^{\\\\sqrt{x}}\\\\:- u^2\\\\texttt{e}^u\\\\:+ 2\\\\int u\\\\texttt{e}^udu<br \\\/>&amp;= x\\\\texttt{e}^{\\\\sqrt{x}}\\\\:- u^2\\\\texttt{e}^u\\\\:+ 2\\\\left( \\\\color{red}{\\\\underbrace{\\\\color{black}{u}}_{\\\\large m}} \\\\color{red}{\\\\underbrace{\\\\color{black}{\\\\texttt{e}^u}}_{\\\\large w}}\\\\:- \\\\int \\\\color{red}{\\\\underbrace{\\\\color{black}{\\\\texttt{e}^u}}_{\\\\large w}} \\\\color{red}{\\\\underbrace{\\\\color{black}{du}}_{dm}} \\\\right)<br \\\/>&amp; \\\\color{gray}{\\\\text{[Apply the substitution]}} \\\\\\\\[5pt]<br \\\/>&amp;= x\\\\texttt{e}^{\\\\sqrt{x}}\\\\:- u^2\\\\texttt{e}^u\\\\:+ 2u\\\\texttt{e}^u\\\\:- 2\\\\int \\\\texttt{e}^udu<br \\\/>&amp; \\\\color{gray}{\\\\text{[Rearrange the expression]}} \\\\\\\\[5pt]<br \\\/>&amp;= x\\\\texttt{e}^{\\\\sqrt{x}}\\\\:- u^2\\\\texttt{e}^u\\\\:+ 2u\\\\texttt{e}^u\\\\:- 2\\\\texttt{e}^u\\\\:+ C<br \\\/>&amp; \\\\color{gray}{\\\\text{[Intergate]}} \\\\\\\\[5pt]<br \\\/>&amp;= x\\\\texttt{e}^{\\\\sqrt{x}}\\\\:- x\\\\texttt{e}^{\\\\sqrt{x}}\\\\:+ 2\\\\sqrt{x}\\\\texttt{e}^{\\\\sqrt{x}}\\\\:- 2\\\\texttt{e}^{\\\\sqrt{x}}\\\\:+ C<br \\\/>&amp; \\\\color{gray}{\\\\text{[Substitute back to x]}} \\\\\\\\[5pt]<br \\\/>&amp;= 2\\\\sqrt{x}\\\\texttt{e}^{\\\\sqrt{x}}\\\\:- 2\\\\texttt{e}^{\\\\sqrt{x}}\\\\:+ C<br \\\/>&amp; \\\\color{gray}{\\\\text{[Simplify]}} \\\\\\\\[5pt]<br \\\/>&amp;= \\\\color{green}{2\\\\texttt{e}^{\\\\sqrt{x}}(\\\\sqrt{x}-1)\\\\:+ C}<br \\\/>&amp; \\\\color{gray}{\\\\text{[Factor]}} \\\\\\\\[5pt]<br \\\/>\\\\end{align*}<br \\\/>\\\\)<p>\\u00a0<\\\/p>\"}}]}<\/script>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t<section class=\"elementor-section elementor-top-section elementor-element elementor-element-f1c9ed4 elementor-section-boxed elementor-section-height-default elementor-section-height-default\" data-id=\"f1c9ed4\" data-element_type=\"section\">\n\t\t\t\t\t\t<div class=\"elementor-container elementor-column-gap-default\">\n\t\t\t\t\t<div class=\"elementor-column elementor-col-100 elementor-top-column elementor-element elementor-element-c41976a\" data-id=\"c41976a\" data-element_type=\"column\">\n\t\t\t<div class=\"elementor-widget-wrap elementor-element-populated\">\n\t\t\t\t\t\t<div class=\"elementor-element elementor-element-762a51f6 elementor-widget elementor-widget-toggle\" data-id=\"762a51f6\" data-element_type=\"widget\" data-widget_type=\"toggle.default\">\n\t\t\t\t<div class=\"elementor-widget-container\">\n\t\t\t\t\t\t\t<div class=\"elementor-toggle\">\n\t\t\t\t\t\t\t<div class=\"elementor-toggle-item\">\n\t\t\t\t\t<h2 id=\"elementor-tab-title-1981\" class=\"elementor-tab-title\" data-tab=\"1\" role=\"button\" aria-controls=\"elementor-tab-content-1981\" aria-expanded=\"false\">\n\t\t\t\t\t\t\t\t\t\t\t\t<span class=\"elementor-toggle-icon elementor-toggle-icon-left\" aria-hidden=\"true\">\n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t<span class=\"elementor-toggle-icon-closed\"><i class=\"fas fa-toggle-off\"><\/i><\/span>\n\t\t\t\t\t\t\t\t<span class=\"elementor-toggle-icon-opened\"><i class=\"elementor-toggle-icon-opened fas fa-toggle-on\"><\/i><\/span>\n\t\t\t\t\t\t\t\t\t\t\t\t\t<\/span>\n\t\t\t\t\t\t\t\t\t\t\t\t<a class=\"elementor-toggle-title\" tabindex=\"0\">Problem 6: Solution<\/a>\n\t\t\t\t\t<\/h2>\n\n\t\t\t\t\t<div id=\"elementor-tab-content-1981\" class=\"elementor-tab-content elementor-clearfix\" data-tab=\"1\" role=\"region\" aria-labelledby=\"elementor-tab-title-1981\"><p><span style=\"font-size: 18.6667px;\">Let \\(u=\\texttt{sin(ln(x))}\\), so \\(du=\\frac{\\texttt{cos(ln(x))}}{x}dx\\); and \\(dv=dx\\), so \\(v=x\\).<\/span><span style=\"font-size: 18.6667px;\">\u00a0<\/span><span style=\"font-size: 18.6667px;\">Substituting these values into \\(uv\\:- \\int vdu\\) leads to the following:<\/span><\/p>\\(\\normalsize<br \/>\\begin{align*}\\displaystyle<br \/>\\int \\texttt{sin(ln(x))}dx<br \/>&amp;= \\color{red}{\\underbrace{\\color{black}{x}}_{\\large v}} \\color{red}{\\underbrace{\\color{black}{\\texttt{sin(ln(x))}}}_{\\large u}}\\:- \\int \\color{red}{\\underbrace{\\color{black}{x}}_{\\large v}} \\color{red}{\\underbrace{\\color{black}{\\left(\u00a0\\frac{\\texttt{cos(ln(x))}}{x}dx \\right)}}_{\\large du}}<br \/>&amp; \\color{gray}{\\text{[Apply the substitution]}} \\\\[5pt]<br \/>&amp;= x\\texttt{sin(ln(x))}\\:- \\int\\texttt{cos(ln(x))} dx<br \/>&amp; \\color{gray}{\\text{[Simplify]}} \\\\[5pt]<br \/>\\end{align*}<br \/>\\)<p>\u00a0<\/p><p><span style=\"font-size: 18.6667px;\">Use IBP again.\u00a0<\/span><span style=\"font-size: 18.6667px;\">Let \\(u=\\texttt{cos(ln(x))}\\), so \\(du=\\frac{-\\texttt{sin(ln(x))}}{x}dx\\); and \\(dv=dx\\), so \\(v=x\\).<\/span><span style=\"font-size: 18.6667px;\">\u00a0<\/span><span style=\"font-size: 18.6667px;\">Substituting these values into \\(uv\\:- \\int vdu\\) leads to the following:<\/span><\/p>\\(\\normalsize<br \/>\\begin{align*}\\displaystyle<br \/>x\\texttt{sin(ln(x))}\\:- \\int\\texttt{cos(ln(x))} dx<br \/>&amp;= x\\texttt{sin(ln(x))}\\:- \\left( \\color{red}{\\underbrace{\\color{black}{x}}_{\\large v}} \\color{red}{\\underbrace{\\color{black}{\\texttt{cos(ln(x))}}}_{\\large u}}\\:- \\int \\color{red}{\\underbrace{\\color{black}{x}}_{\\large v}} \\color{red}{\\underbrace{\\color{black}{\\left(\u00a0\\frac{-\\texttt{sin(ln(x))}}{x}dx \\right)}}_{\\large du}} \\right)<br \/>&amp; \\color{gray}{\\text{[Apply the substitution]}} \\\\[5pt]<br \/>&amp;= x\\texttt{sin(ln(x))}\\:- x\\texttt{cos(ln(x))}\\:- \\int \\texttt{sin(ln(x))}dx<br \/>&amp; \\color{gray}{\\text{[Simplify]}}<br \/>\\end{align*}<br \/>\\)<p>\u00a0<\/p><p><span style=\"font-size: 18.6667px;\">Given that we ended up with the same integral we wanted to solve for, we can equate the original integrand to the resulting expression and perform algebraic manipulations while treating the reappearing integral as a variable to solve for:<\/span><\/p>\\(\\normalsize<br \/>\\begin{align*}\\displaystyle<br \/>&amp; \\color{red}{\\int \\texttt{sin(ln(x))}dx} = x\\texttt{sin(ln(x))}\\:- x\\texttt{cos(ln(x))}\\:- \\color{red}{\\int \\texttt{sin(ln(x))}dx}<br \/>&amp; \\color{gray}{\\text{[Form an equation with both integrals]}} \\\\[5pt]<br \/>&amp; \\color{red}{2\\int \\texttt{sin(ln(x))}dx} = x\\texttt{sin(ln(x))}\\:- x\\texttt{cos(ln(x))}\\:+ C<br \/>&amp; \\color{gray}{\\text{[Arrange both integrals to the left hand side of the equation]}} \\\\[5pt]<br \/>&amp; \\color{green}{\\int \\texttt{sin(ln(x))}dx = \\frac{x}{2} [\\texttt{sin(ln(x))}\\:- \\texttt{cos(ln(x))}]\\:+ C}<br \/>&amp; \\color{gray}{\\text{[Divide both sides by 2 and factor x]}}<br \/>\\end{align*}<br \/>\\)<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t\t\t\t\t<script type=\"application\/ld+json\">{\"@context\":\"https:\\\/\\\/schema.org\",\"@type\":\"FAQPage\",\"mainEntity\":[{\"@type\":\"Question\",\"name\":\"Problem 6: Solution\",\"acceptedAnswer\":{\"@type\":\"Answer\",\"text\":\"<p><span style=\\\"font-size: 18.6667px;\\\">Let \\\\(u=\\\\texttt{sin(ln(x))}\\\\), so \\\\(du=\\\\frac{\\\\texttt{cos(ln(x))}}{x}dx\\\\); and \\\\(dv=dx\\\\), so \\\\(v=x\\\\).<\\\/span><span style=\\\"font-size: 18.6667px;\\\">\\u00a0<\\\/span><span style=\\\"font-size: 18.6667px;\\\">Substituting these values into \\\\(uv\\\\:- \\\\int vdu\\\\) leads to the following:<\\\/span><\\\/p>\\\\(\\\\normalsize<br \\\/>\\\\begin{align*}\\\\displaystyle<br \\\/>\\\\int \\\\texttt{sin(ln(x))}dx<br \\\/>&amp;= \\\\color{red}{\\\\underbrace{\\\\color{black}{x}}_{\\\\large v}} \\\\color{red}{\\\\underbrace{\\\\color{black}{\\\\texttt{sin(ln(x))}}}_{\\\\large u}}\\\\:- \\\\int \\\\color{red}{\\\\underbrace{\\\\color{black}{x}}_{\\\\large v}} \\\\color{red}{\\\\underbrace{\\\\color{black}{\\\\left(\\u00a0\\\\frac{\\\\texttt{cos(ln(x))}}{x}dx \\\\right)}}_{\\\\large du}}<br \\\/>&amp; \\\\color{gray}{\\\\text{[Apply the substitution]}} \\\\\\\\[5pt]<br \\\/>&amp;= x\\\\texttt{sin(ln(x))}\\\\:- \\\\int\\\\texttt{cos(ln(x))} dx<br \\\/>&amp; \\\\color{gray}{\\\\text{[Simplify]}} \\\\\\\\[5pt]<br \\\/>\\\\end{align*}<br \\\/>\\\\)<p>\\u00a0<\\\/p><p><span style=\\\"font-size: 18.6667px;\\\">Use IBP again.\\u00a0<\\\/span><span style=\\\"font-size: 18.6667px;\\\">Let \\\\(u=\\\\texttt{cos(ln(x))}\\\\), so \\\\(du=\\\\frac{-\\\\texttt{sin(ln(x))}}{x}dx\\\\); and \\\\(dv=dx\\\\), so \\\\(v=x\\\\).<\\\/span><span style=\\\"font-size: 18.6667px;\\\">\\u00a0<\\\/span><span style=\\\"font-size: 18.6667px;\\\">Substituting these values into \\\\(uv\\\\:- \\\\int vdu\\\\) leads to the following:<\\\/span><\\\/p>\\\\(\\\\normalsize<br \\\/>\\\\begin{align*}\\\\displaystyle<br \\\/>x\\\\texttt{sin(ln(x))}\\\\:- \\\\int\\\\texttt{cos(ln(x))} dx<br \\\/>&amp;= x\\\\texttt{sin(ln(x))}\\\\:- \\\\left( \\\\color{red}{\\\\underbrace{\\\\color{black}{x}}_{\\\\large v}} \\\\color{red}{\\\\underbrace{\\\\color{black}{\\\\texttt{cos(ln(x))}}}_{\\\\large u}}\\\\:- \\\\int \\\\color{red}{\\\\underbrace{\\\\color{black}{x}}_{\\\\large v}} \\\\color{red}{\\\\underbrace{\\\\color{black}{\\\\left(\\u00a0\\\\frac{-\\\\texttt{sin(ln(x))}}{x}dx \\\\right)}}_{\\\\large du}} \\\\right)<br \\\/>&amp; \\\\color{gray}{\\\\text{[Apply the substitution]}} \\\\\\\\[5pt]<br \\\/>&amp;= x\\\\texttt{sin(ln(x))}\\\\:- x\\\\texttt{cos(ln(x))}\\\\:- \\\\int \\\\texttt{sin(ln(x))}dx<br \\\/>&amp; \\\\color{gray}{\\\\text{[Simplify]}}<br \\\/>\\\\end{align*}<br \\\/>\\\\)<p>\\u00a0<\\\/p><p><span style=\\\"font-size: 18.6667px;\\\">Given that we ended up with the same integral we wanted to solve for, we can equate the original integrand to the resulting expression and perform algebraic manipulations while treating the reappearing integral as a variable to solve for:<\\\/span><\\\/p>\\\\(\\\\normalsize<br \\\/>\\\\begin{align*}\\\\displaystyle<br \\\/>&amp; \\\\color{red}{\\\\int \\\\texttt{sin(ln(x))}dx} = x\\\\texttt{sin(ln(x))}\\\\:- x\\\\texttt{cos(ln(x))}\\\\:- \\\\color{red}{\\\\int \\\\texttt{sin(ln(x))}dx}<br \\\/>&amp; \\\\color{gray}{\\\\text{[Form an equation with both integrals]}} \\\\\\\\[5pt]<br \\\/>&amp; \\\\color{red}{2\\\\int \\\\texttt{sin(ln(x))}dx} = x\\\\texttt{sin(ln(x))}\\\\:- x\\\\texttt{cos(ln(x))}\\\\:+ C<br \\\/>&amp; \\\\color{gray}{\\\\text{[Arrange both integrals to the left hand side of the equation]}} \\\\\\\\[5pt]<br \\\/>&amp; \\\\color{green}{\\\\int \\\\texttt{sin(ln(x))}dx = \\\\frac{x}{2} [\\\\texttt{sin(ln(x))}\\\\:- \\\\texttt{cos(ln(x))}]\\\\:+ C}<br \\\/>&amp; \\\\color{gray}{\\\\text{[Divide both sides by 2 and factor x]}}<br \\\/>\\\\end{align*}<br \\\/>\\\\)\"}}]}<\/script>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t<section class=\"elementor-section elementor-top-section elementor-element elementor-element-67d8036e elementor-section-boxed elementor-section-height-default elementor-section-height-default\" data-id=\"67d8036e\" data-element_type=\"section\">\n\t\t\t\t\t\t<div class=\"elementor-container elementor-column-gap-default\">\n\t\t\t\t\t<div class=\"elementor-column elementor-col-100 elementor-top-column elementor-element elementor-element-19ab69b0\" data-id=\"19ab69b0\" data-element_type=\"column\">\n\t\t\t<div class=\"elementor-widget-wrap elementor-element-populated\">\n\t\t\t\t\t\t<div class=\"elementor-element elementor-element-636b8d40 elementor-widget elementor-widget-toggle\" data-id=\"636b8d40\" data-element_type=\"widget\" data-widget_type=\"toggle.default\">\n\t\t\t\t<div class=\"elementor-widget-container\">\n\t\t\t\t\t\t\t<div class=\"elementor-toggle\">\n\t\t\t\t\t\t\t<div class=\"elementor-toggle-item\">\n\t\t\t\t\t<h2 id=\"elementor-tab-title-1661\" class=\"elementor-tab-title\" data-tab=\"1\" role=\"button\" aria-controls=\"elementor-tab-content-1661\" aria-expanded=\"false\">\n\t\t\t\t\t\t\t\t\t\t\t\t<span class=\"elementor-toggle-icon elementor-toggle-icon-left\" aria-hidden=\"true\">\n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t<span class=\"elementor-toggle-icon-closed\"><i class=\"fas fa-toggle-off\"><\/i><\/span>\n\t\t\t\t\t\t\t\t<span class=\"elementor-toggle-icon-opened\"><i class=\"elementor-toggle-icon-opened fas fa-toggle-on\"><\/i><\/span>\n\t\t\t\t\t\t\t\t\t\t\t\t\t<\/span>\n\t\t\t\t\t\t\t\t\t\t\t\t<a class=\"elementor-toggle-title\" tabindex=\"0\">Problem 7: Solution<\/a>\n\t\t\t\t\t<\/h2>\n\n\t\t\t\t\t<div id=\"elementor-tab-content-1661\" class=\"elementor-tab-content elementor-clearfix\" data-tab=\"1\" role=\"region\" aria-labelledby=\"elementor-tab-title-1661\"><p><span style=\"font-size: 18.6667px;\">Let \\(u=\\texttt{cos}(5x)\\), so \\(du=-5\\texttt{sin}(5x)dx\\); and \\(dv=\\texttt{e}^{2x}dx\\), so \\(v=\\frac{1}{2}\\texttt{e}^{2x}\\).<\/span><span style=\"font-size: 18.6667px;\">\u00a0<\/span><span style=\"font-size: 18.6667px;\">Substituting these values into \\(uv\\:- \\int vdu\\) leads to the following:<\/span><\/p>\\(\\normalsize<br \/>\\begin{align*}\\displaystyle<br \/>\\int \\texttt{e}^{2x}\\texttt{cos}(5x)dx<br \/>&amp;= \\color{red}{\\underbrace{\\color{black}{\\frac{1}{2}\\texttt{e}^{2x}}}_{\\large v}} \\color{red}{\\underbrace{\\color{black}{\\texttt{cos}(5x)}}_{\\large u}}\\:- \\int \\color{red}{\\underbrace{\\color{black}{\\left( \\frac{1}{2}\\texttt{e}^{2x} \\right)}}_{\\large v}} \\color{red}{\\underbrace{\\color{black}{(-5\\texttt{cos}(5x)dx)}}_{\\large du}}<br \/>&amp; \\color{gray}{\\text{[Apply the substitution]}} \\\\[5pt]<br \/>&amp;= \\frac{1}{2}\\texttt{e}^{2x}\\texttt{cos}(5x)\\:+ \\frac{5}{2}\\int \\texttt{e}^{2x}\\texttt{sin}(5x)dx<br \/>&amp; \\color{gray}{\\text{[Simplify]}} \\\\[5pt]<br \/>\\end{align*}<br \/>\\)<p>\u00a0<\/p><p>\u00a0<\/p><p><span style=\"font-size: 18.6667px;\">Use IBP again.\u00a0<\/span><span style=\"font-size: 18.6667px;\">Let \\(u=\\texttt{sin}(5x)\\), so \\(du=5\\texttt{cos}(5x)dx\\); and \\(dv=\\texttt{e}^{2x}dx\\), so \\(v=\\frac{1}{2}\\texttt{e}^{2x}\\).<\/span><span style=\"font-size: 18.6667px;\">\u00a0<\/span><span style=\"font-size: 18.6667px;\">Substituting these values into \\(uv\\:- \\int vdu\\) leads to the following:<\/span><\/p>\\(\\normalsize<br \/>\\begin{align*}\\displaystyle<br \/>\\frac{1}{2}\\texttt{e}^{2x}\\texttt{cos}(5x)\\:+ \\frac{5}{2}\\int \\texttt{e}^{2x}\\texttt{sin}(5x)dx<br \/>&amp;=\\frac{1}{2}\\texttt{e}^{2x}\\texttt{cos}(5x)\\:+ \\frac{5}{2}\u00a0\\left( \\color{red}{\\underbrace{\\color{black}{\\frac{1}{2}\\texttt{e}^{2x}}}_{\\large v}} \\color{red}{\\underbrace{\\color{black}{\\texttt{sin}(5x)}}_{\\large u}}\\:- \\int \\color{red}{\\underbrace{\\color{black}{\\left( \\frac{1}{2}\\texttt{e}^{2x} \\right)}}_{\\large v}} \\color{red}{\\underbrace{\\color{black}{(5\\texttt{cos}(5x)dx)}}_{\\large du}}\u00a0\\right)<br \/>&amp; \\color{gray}{\\text{[Apply the substitution]}} \\\\[5pt]<br \/>&amp;= \\frac{1}{2} \\texttt{e}^{2x}\\texttt{cos}(5x)\\:+ \\frac{5}{4}\\texttt{e}^{2x}\\texttt{sin}(5x)\\:- \\frac{25}{4}\\int \\texttt{e}^{2x}\\texttt{cos}(5x)dx<br \/>&amp; \\color{gray}{\\text{[Simplify]}}<br \/>\\end{align*}<br \/>\\)<p>\u00a0<\/p><p>\u00a0<\/p><p><span style=\"font-size: 18.6667px;\">Given that we ended up with the same integral we wanted to solve for, we can equate the original integrand to the resulting expression and perform algebraic manipulations while treating the reappearing integral as a variable to solve for:<\/span><\/p>\\(\\normalsize<br \/>\\begin{align*}\\displaystyle<br \/>&amp; \\color{red}{\\int \\texttt{e}^{2x}\\texttt{cos}(5x)dx} = \\frac{1}{2} \\texttt{e}^{2x}\\texttt{cos}(5x)\\:+ \\frac{5}{4}\\texttt{e}^{2x}\\texttt{sin}(5x)\\:- \\color{red}{\\frac{25}{4}\\int \\texttt{e}^{2x}\\texttt{cos}(5x)dx}<br \/>&amp; \\color{gray}{\\text{[Form an equation with both integrals]}} \\\\[5pt]<br \/>&amp; \\color{red}{\\frac{29}{4}\\int \\texttt{e}^{2x}\\texttt{cos}(5x)dx} = \\frac{1}{2} \\texttt{e}^{2x}\\texttt{cos}(5x)\\:+ \\frac{5}{4}\\texttt{e}^{2x}\\texttt{sin}(5x)\\:+ C<br \/>&amp; \\color{gray}{\\text{[Arrange both integrals to the left hand side of the equation]}} \\\\[5pt]<br \/>&amp; \\color{red}{\\int \\texttt{e}^{2x}\\texttt{cos}(5x)dx} = \\left( \\frac{4}{29} \\right) \\left( \\frac{1}{2}\\texttt{e}^{2x} \\right) (\\texttt{cos}(5x)\\:+ \\frac{5}{2}\\texttt{sin}(5x))\\:+ C<br \/>&amp; \\color{gray}{\\text{[Factor]}} \\\\[5pt]<br \/>&amp; \\color{green}{\\int \\texttt{e}^{2x}\\texttt{cos}(5x)dx = \\left( \\frac{1}{29}\\texttt{e}^{2x} \\right)[2\\texttt{cos}(5x)\\:+ 5\\texttt{sin}(5x)]\\:+ C}<br \/>&amp; \\color{gray}{\\text{[Simplify and distribute the numerator]}}<br \/>\\end{align*}<br \/>\\)<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t\t\t\t\t<script type=\"application\/ld+json\">{\"@context\":\"https:\\\/\\\/schema.org\",\"@type\":\"FAQPage\",\"mainEntity\":[{\"@type\":\"Question\",\"name\":\"Problem 7: Solution\",\"acceptedAnswer\":{\"@type\":\"Answer\",\"text\":\"<p><span style=\\\"font-size: 18.6667px;\\\">Let \\\\(u=\\\\texttt{cos}(5x)\\\\), so \\\\(du=-5\\\\texttt{sin}(5x)dx\\\\); and \\\\(dv=\\\\texttt{e}^{2x}dx\\\\), so \\\\(v=\\\\frac{1}{2}\\\\texttt{e}^{2x}\\\\).<\\\/span><span style=\\\"font-size: 18.6667px;\\\">\\u00a0<\\\/span><span style=\\\"font-size: 18.6667px;\\\">Substituting these values into \\\\(uv\\\\:- \\\\int vdu\\\\) leads to the following:<\\\/span><\\\/p>\\\\(\\\\normalsize<br \\\/>\\\\begin{align*}\\\\displaystyle<br \\\/>\\\\int \\\\texttt{e}^{2x}\\\\texttt{cos}(5x)dx<br \\\/>&amp;= \\\\color{red}{\\\\underbrace{\\\\color{black}{\\\\frac{1}{2}\\\\texttt{e}^{2x}}}_{\\\\large v}} \\\\color{red}{\\\\underbrace{\\\\color{black}{\\\\texttt{cos}(5x)}}_{\\\\large u}}\\\\:- \\\\int \\\\color{red}{\\\\underbrace{\\\\color{black}{\\\\left( \\\\frac{1}{2}\\\\texttt{e}^{2x} \\\\right)}}_{\\\\large v}} \\\\color{red}{\\\\underbrace{\\\\color{black}{(-5\\\\texttt{cos}(5x)dx)}}_{\\\\large du}}<br \\\/>&amp; \\\\color{gray}{\\\\text{[Apply the substitution]}} \\\\\\\\[5pt]<br \\\/>&amp;= \\\\frac{1}{2}\\\\texttt{e}^{2x}\\\\texttt{cos}(5x)\\\\:+ \\\\frac{5}{2}\\\\int \\\\texttt{e}^{2x}\\\\texttt{sin}(5x)dx<br \\\/>&amp; \\\\color{gray}{\\\\text{[Simplify]}} \\\\\\\\[5pt]<br \\\/>\\\\end{align*}<br \\\/>\\\\)<p>\\u00a0<\\\/p><p>\\u00a0<\\\/p><p><span style=\\\"font-size: 18.6667px;\\\">Use IBP again.\\u00a0<\\\/span><span style=\\\"font-size: 18.6667px;\\\">Let \\\\(u=\\\\texttt{sin}(5x)\\\\), so \\\\(du=5\\\\texttt{cos}(5x)dx\\\\); and \\\\(dv=\\\\texttt{e}^{2x}dx\\\\), so \\\\(v=\\\\frac{1}{2}\\\\texttt{e}^{2x}\\\\).<\\\/span><span style=\\\"font-size: 18.6667px;\\\">\\u00a0<\\\/span><span style=\\\"font-size: 18.6667px;\\\">Substituting these values into \\\\(uv\\\\:- \\\\int vdu\\\\) leads to the following:<\\\/span><\\\/p>\\\\(\\\\normalsize<br \\\/>\\\\begin{align*}\\\\displaystyle<br \\\/>\\\\frac{1}{2}\\\\texttt{e}^{2x}\\\\texttt{cos}(5x)\\\\:+ \\\\frac{5}{2}\\\\int \\\\texttt{e}^{2x}\\\\texttt{sin}(5x)dx<br \\\/>&amp;=\\\\frac{1}{2}\\\\texttt{e}^{2x}\\\\texttt{cos}(5x)\\\\:+ \\\\frac{5}{2}\\u00a0\\\\left( \\\\color{red}{\\\\underbrace{\\\\color{black}{\\\\frac{1}{2}\\\\texttt{e}^{2x}}}_{\\\\large v}} \\\\color{red}{\\\\underbrace{\\\\color{black}{\\\\texttt{sin}(5x)}}_{\\\\large u}}\\\\:- \\\\int \\\\color{red}{\\\\underbrace{\\\\color{black}{\\\\left( \\\\frac{1}{2}\\\\texttt{e}^{2x} \\\\right)}}_{\\\\large v}} \\\\color{red}{\\\\underbrace{\\\\color{black}{(5\\\\texttt{cos}(5x)dx)}}_{\\\\large du}}\\u00a0\\\\right)<br \\\/>&amp; \\\\color{gray}{\\\\text{[Apply the substitution]}} \\\\\\\\[5pt]<br \\\/>&amp;= \\\\frac{1}{2} \\\\texttt{e}^{2x}\\\\texttt{cos}(5x)\\\\:+ \\\\frac{5}{4}\\\\texttt{e}^{2x}\\\\texttt{sin}(5x)\\\\:- \\\\frac{25}{4}\\\\int \\\\texttt{e}^{2x}\\\\texttt{cos}(5x)dx<br \\\/>&amp; \\\\color{gray}{\\\\text{[Simplify]}}<br \\\/>\\\\end{align*}<br \\\/>\\\\)<p>\\u00a0<\\\/p><p>\\u00a0<\\\/p><p><span style=\\\"font-size: 18.6667px;\\\">Given that we ended up with the same integral we wanted to solve for, we can equate the original integrand to the resulting expression and perform algebraic manipulations while treating the reappearing integral as a variable to solve for:<\\\/span><\\\/p>\\\\(\\\\normalsize<br \\\/>\\\\begin{align*}\\\\displaystyle<br \\\/>&amp; \\\\color{red}{\\\\int \\\\texttt{e}^{2x}\\\\texttt{cos}(5x)dx} = \\\\frac{1}{2} \\\\texttt{e}^{2x}\\\\texttt{cos}(5x)\\\\:+ \\\\frac{5}{4}\\\\texttt{e}^{2x}\\\\texttt{sin}(5x)\\\\:- \\\\color{red}{\\\\frac{25}{4}\\\\int \\\\texttt{e}^{2x}\\\\texttt{cos}(5x)dx}<br \\\/>&amp; \\\\color{gray}{\\\\text{[Form an equation with both integrals]}} \\\\\\\\[5pt]<br \\\/>&amp; \\\\color{red}{\\\\frac{29}{4}\\\\int \\\\texttt{e}^{2x}\\\\texttt{cos}(5x)dx} = \\\\frac{1}{2} \\\\texttt{e}^{2x}\\\\texttt{cos}(5x)\\\\:+ \\\\frac{5}{4}\\\\texttt{e}^{2x}\\\\texttt{sin}(5x)\\\\:+ C<br \\\/>&amp; \\\\color{gray}{\\\\text{[Arrange both integrals to the left hand side of the equation]}} \\\\\\\\[5pt]<br \\\/>&amp; \\\\color{red}{\\\\int \\\\texttt{e}^{2x}\\\\texttt{cos}(5x)dx} = \\\\left( \\\\frac{4}{29} \\\\right) \\\\left( \\\\frac{1}{2}\\\\texttt{e}^{2x} \\\\right) (\\\\texttt{cos}(5x)\\\\:+ \\\\frac{5}{2}\\\\texttt{sin}(5x))\\\\:+ C<br \\\/>&amp; \\\\color{gray}{\\\\text{[Factor]}} \\\\\\\\[5pt]<br \\\/>&amp; \\\\color{green}{\\\\int \\\\texttt{e}^{2x}\\\\texttt{cos}(5x)dx = \\\\left( \\\\frac{1}{29}\\\\texttt{e}^{2x} \\\\right)[2\\\\texttt{cos}(5x)\\\\:+ 5\\\\texttt{sin}(5x)]\\\\:+ C}<br \\\/>&amp; \\\\color{gray}{\\\\text{[Simplify and distribute the numerator]}}<br \\\/>\\\\end{align*}<br \\\/>\\\\)\"}}]}<\/script>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t<section class=\"elementor-section elementor-top-section elementor-element elementor-element-389ae58d elementor-section-boxed elementor-section-height-default elementor-section-height-default\" data-id=\"389ae58d\" data-element_type=\"section\">\n\t\t\t\t\t\t<div class=\"elementor-container elementor-column-gap-default\">\n\t\t\t\t\t<div class=\"elementor-column elementor-col-100 elementor-top-column elementor-element elementor-element-4944ef71\" data-id=\"4944ef71\" data-element_type=\"column\">\n\t\t\t<div class=\"elementor-widget-wrap elementor-element-populated\">\n\t\t\t\t\t\t<div class=\"elementor-element elementor-element-41c1a588 elementor-widget elementor-widget-toggle\" data-id=\"41c1a588\" data-element_type=\"widget\" data-widget_type=\"toggle.default\">\n\t\t\t\t<div class=\"elementor-widget-container\">\n\t\t\t\t\t\t\t<div class=\"elementor-toggle\">\n\t\t\t\t\t\t\t<div class=\"elementor-toggle-item\">\n\t\t\t\t\t<h2 id=\"elementor-tab-title-1101\" class=\"elementor-tab-title\" data-tab=\"1\" role=\"button\" aria-controls=\"elementor-tab-content-1101\" aria-expanded=\"false\">\n\t\t\t\t\t\t\t\t\t\t\t\t<span class=\"elementor-toggle-icon elementor-toggle-icon-left\" aria-hidden=\"true\">\n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t<span class=\"elementor-toggle-icon-closed\"><i class=\"fas fa-toggle-off\"><\/i><\/span>\n\t\t\t\t\t\t\t\t<span class=\"elementor-toggle-icon-opened\"><i class=\"elementor-toggle-icon-opened fas fa-toggle-on\"><\/i><\/span>\n\t\t\t\t\t\t\t\t\t\t\t\t\t<\/span>\n\t\t\t\t\t\t\t\t\t\t\t\t<a class=\"elementor-toggle-title\" tabindex=\"0\">Problem 8: Solution<\/a>\n\t\t\t\t\t<\/h2>\n\n\t\t\t\t\t<div id=\"elementor-tab-content-1101\" class=\"elementor-tab-content elementor-clearfix\" data-tab=\"1\" role=\"region\" aria-labelledby=\"elementor-tab-title-1101\"><p><span style=\"font-size: 18.6667px;\">Let \\(u=\\texttt{sin}(\\beta x)\\), so \\(du=\\beta \\texttt{cos}(\\beta x)dx\\); and \\(dv=\\texttt{e}^{\\alpha x}dx\\), so \\(v=\\frac{1}{\\alpha}\\texttt{e}^{\\alpha x}\\).<\/span><span style=\"font-size: 18.6667px;\">\u00a0<\/span><span style=\"font-size: 18.6667px;\">Substituting these values into \\(uv\\:- \\int vdu\\) leads to the following:<\/span><\/p>\\(\\normalsize<br \/>\\begin{align*}\\displaystyle<br \/>\\int \\texttt{e}^{\\alpha x}\\texttt{sin}(\\beta x)dx<br \/>&amp;= \\color{red}{\\underbrace{\\color{black}{\\frac{1}{\\alpha}\\texttt{e}^{\\alpha x}}}_{\\large v}} \\color{red}{\\underbrace{\\color{black}{\\texttt{sin}(\\beta x)}}_{\\large u}}\\:- \\int \\color{red}{\\underbrace{\\color{black}{\\left( \\frac{1}{\\alpha}\\texttt{e}^{\\alpha x} \\right)}}_{\\large v}} \\color{red}{\\underbrace{\\color{black}{(\\beta \\texttt{cos}(\\beta x)dx)}}_{\\large du}}<br \/>&amp; \\color{gray}{\\text{[Apply the substitution]}} \\\\[5pt]<br \/>&amp;= \\frac{1}{\\alpha}\\texttt{e}^{\\alpha x}\\texttt{sin}(\\beta x)\\:- \\frac{\\beta}{\\alpha}\\int \\texttt{e}^{\\alpha x}\\texttt{cos}(\\beta x)dx<br \/>&amp; \\color{gray}{\\text{[Simplify]}} \\\\[5pt]<br \/>\\end{align*}<br \/>\\)<p>\u00a0<\/p><p><span style=\"font-size: 18.6667px;\">Use IBP again.\u00a0<\/span><span style=\"font-size: 18.6667px;\">Let \\(u=\\texttt{cos}(\\beta x)\\), so \\(du=-\\beta \\texttt{sin}(\\beta x)dx\\); and \\(dv=\\texttt{e}^{\\alpha x}dx\\), so \\(v=\\frac{1}{\\alpha }\\texttt{e}^{\\alpha x}\\).<\/span><span style=\"font-size: 18.6667px;\">\u00a0<\/span><span style=\"font-size: 18.6667px;\">Substituting these values into \\(uv\\:- \\int vdu\\) leads to the following:<\/span><\/p>\\(\\normalsize<br \/>\\begin{align*}\\displaystyle<br \/>\\frac{1}{\\alpha}\\texttt{e}^{\\alpha x}\\texttt{sin}(\\beta x)\\:- \\frac{\\beta}{\\alpha}\\int \\texttt{e}^{\\alpha x}\\texttt{cos}(\\beta x)dx<br \/>&amp;=\\frac{1}{\\alpha}\\texttt{e}^{\\alpha x}\\texttt{sin}(\\beta x)\\:- \\frac{\\beta}{\\alpha}\u00a0\\left( \\color{red}{\\underbrace{\\color{black}{\\frac{1}{\\alpha}\\texttt{e}^{\\alpha x}}}_{\\large v}} \\color{red}{\\underbrace{\\color{black}{\\texttt{cos}(\\beta x)}}_{\\large u}}\\:- \\int \\color{red}{\\underbrace{\\color{black}{\\left( \\frac{1}{\\alpha}\\texttt{e}^{\\alpha x} \\right)}}_{\\large v}} \\color{red}{\\underbrace{\\color{black}{(-\\beta \\texttt{sin}(\\beta x)dx)}}_{\\large du}}\u00a0\\right)<br \/>&amp; \\color{gray}{\\text{[Apply the substitution]}} \\\\[5pt]<br \/>&amp;= \\frac{1}{\\alpha} \\texttt{e}^{\\alpha x}\\texttt{sin}(\\beta x)\\:- \\frac{\\beta}{\\alpha^2}\\texttt{e}^{\\alpha x}\\texttt{cos}(\\beta x)\\:- \\frac{\\beta^2}{\\alpha^2}\\int \\texttt{e}^{\\alpha x}\\texttt{sin}(\\beta x)dx<br \/>&amp; \\color{gray}{\\text{[Simplify]}}<br \/>\\end{align*}<br \/>\\)<p>\u00a0<\/p><p><span style=\"font-size: 18.6667px;\">Given that we ended up with the same integral we wanted to solve for, we can equate the original integrand to the resulting expression and perform algebraic manipulations while treating the reappearing integral as a variable to solve for:<\/span><\/p>\\(\\normalsize<br \/>\\begin{align*}\\displaystyle<br \/>&amp; \\color{red}{\\int \\texttt{e}^{\\alpha x}\\texttt{sin}(\\beta x)dx} = \\frac{1}{\\alpha} \\texttt{e}^{\\alpha x}\\texttt{sin}(\\beta x)\\:- \\frac{\\beta}{\\alpha^2}\\texttt{e}^{\\alpha x}\\texttt{cos}(\\beta x)\\:- \\color{red}{\\frac{\\beta^2}{\\alpha^2}\\int \\texttt{e}^{\\alpha x}\\texttt{sin}(\\beta x)dx}<br \/>&amp; \\color{gray}{\\text{[Form an equation with both integrals]}} \\\\[5pt]<br \/>&amp; \\color{red}{\\frac{\\beta^2\\:+ \\alpha^2}{\\alpha^2} \\int \\texttt{e}^{\\alpha x}\\texttt{sin}(\\beta x)dx} = \\frac{1}{\\alpha} \\texttt{e}^{\\alpha x}\\texttt{sin}(\\beta x)\\:- \\frac{\\beta}{\\alpha^2}\\texttt{e}^{\\alpha x}\\texttt{cos}(\\beta x)\\:+ C<br \/>&amp; \\color{gray}{\\text{[Arrange both integrals to the left hand side of the equation]}} \\\\[5pt]<br \/>&amp; \\color{red}{\\int \\texttt{e}^{\\alpha x}\\texttt{sin}(\\beta x)dx} = \\left( \\frac{\\alpha^2}{\\alpha^2+\\beta^2} \\right) \\left( \\frac{1}{\\alpha}\\texttt{e}^{\\alpha x} \\right) [\\texttt{sin}(\\beta x)\\:- \\frac{\\beta}{\\alpha}\\texttt{cos}(\\beta x)]\\:+ C<br \/>&amp; \\color{gray}{\\text{[Factor]}} \\\\[5pt]<br \/>&amp; \\int \\texttt{e}^{\\alpha x}\\texttt{sin}(\\beta x)dx = \\left( \\frac{\\alpha}{\\alpha^2+\\beta^2} \\right) \\left(\\texttt{e}^{\\alpha x} \\right) [\\texttt{sin}(\\beta x)\\:- \\frac{\\beta}{\\alpha}\\texttt{cos}(\\beta x)]\\:+ C<br \/>&amp; \\color{gray}{\\text{[Simplify]}} \\\\[5pt]<br \/>&amp; \\color{green}{\\int \\texttt{e}^{\\alpha x}\\texttt{sin}(\\beta x)dx = \\left( \\frac{\\texttt{e}^{\\alpha x}}{\\alpha^2\\:+ \\beta^2} \\right)[\\alpha\\texttt{sin}(\\beta x)\\:- \\beta\\texttt{cos}(\\beta x)]\\:+ C}<br \/>&amp; \\color{gray}{\\text{[Distribute the numerator]}}<br \/>\\end{align*}<br \/>\\)<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t\t\t\t\t<script type=\"application\/ld+json\">{\"@context\":\"https:\\\/\\\/schema.org\",\"@type\":\"FAQPage\",\"mainEntity\":[{\"@type\":\"Question\",\"name\":\"Problem 8: Solution\",\"acceptedAnswer\":{\"@type\":\"Answer\",\"text\":\"<p><span style=\\\"font-size: 18.6667px;\\\">Let \\\\(u=\\\\texttt{sin}(\\\\beta x)\\\\), so \\\\(du=\\\\beta \\\\texttt{cos}(\\\\beta x)dx\\\\); and \\\\(dv=\\\\texttt{e}^{\\\\alpha x}dx\\\\), so \\\\(v=\\\\frac{1}{\\\\alpha}\\\\texttt{e}^{\\\\alpha x}\\\\).<\\\/span><span style=\\\"font-size: 18.6667px;\\\">\\u00a0<\\\/span><span style=\\\"font-size: 18.6667px;\\\">Substituting these values into \\\\(uv\\\\:- \\\\int vdu\\\\) leads to the following:<\\\/span><\\\/p>\\\\(\\\\normalsize<br \\\/>\\\\begin{align*}\\\\displaystyle<br \\\/>\\\\int \\\\texttt{e}^{\\\\alpha x}\\\\texttt{sin}(\\\\beta x)dx<br \\\/>&amp;= \\\\color{red}{\\\\underbrace{\\\\color{black}{\\\\frac{1}{\\\\alpha}\\\\texttt{e}^{\\\\alpha x}}}_{\\\\large v}} \\\\color{red}{\\\\underbrace{\\\\color{black}{\\\\texttt{sin}(\\\\beta x)}}_{\\\\large u}}\\\\:- \\\\int \\\\color{red}{\\\\underbrace{\\\\color{black}{\\\\left( \\\\frac{1}{\\\\alpha}\\\\texttt{e}^{\\\\alpha x} \\\\right)}}_{\\\\large v}} \\\\color{red}{\\\\underbrace{\\\\color{black}{(\\\\beta \\\\texttt{cos}(\\\\beta x)dx)}}_{\\\\large du}}<br \\\/>&amp; \\\\color{gray}{\\\\text{[Apply the substitution]}} \\\\\\\\[5pt]<br \\\/>&amp;= \\\\frac{1}{\\\\alpha}\\\\texttt{e}^{\\\\alpha x}\\\\texttt{sin}(\\\\beta x)\\\\:- \\\\frac{\\\\beta}{\\\\alpha}\\\\int \\\\texttt{e}^{\\\\alpha x}\\\\texttt{cos}(\\\\beta x)dx<br \\\/>&amp; \\\\color{gray}{\\\\text{[Simplify]}} \\\\\\\\[5pt]<br \\\/>\\\\end{align*}<br \\\/>\\\\)<p>\\u00a0<\\\/p><p><span style=\\\"font-size: 18.6667px;\\\">Use IBP again.\\u00a0<\\\/span><span style=\\\"font-size: 18.6667px;\\\">Let \\\\(u=\\\\texttt{cos}(\\\\beta x)\\\\), so \\\\(du=-\\\\beta \\\\texttt{sin}(\\\\beta x)dx\\\\); and \\\\(dv=\\\\texttt{e}^{\\\\alpha x}dx\\\\), so \\\\(v=\\\\frac{1}{\\\\alpha }\\\\texttt{e}^{\\\\alpha x}\\\\).<\\\/span><span style=\\\"font-size: 18.6667px;\\\">\\u00a0<\\\/span><span style=\\\"font-size: 18.6667px;\\\">Substituting these values into \\\\(uv\\\\:- \\\\int vdu\\\\) leads to the following:<\\\/span><\\\/p>\\\\(\\\\normalsize<br \\\/>\\\\begin{align*}\\\\displaystyle<br \\\/>\\\\frac{1}{\\\\alpha}\\\\texttt{e}^{\\\\alpha x}\\\\texttt{sin}(\\\\beta x)\\\\:- \\\\frac{\\\\beta}{\\\\alpha}\\\\int \\\\texttt{e}^{\\\\alpha x}\\\\texttt{cos}(\\\\beta x)dx<br \\\/>&amp;=\\\\frac{1}{\\\\alpha}\\\\texttt{e}^{\\\\alpha x}\\\\texttt{sin}(\\\\beta x)\\\\:- \\\\frac{\\\\beta}{\\\\alpha}\\u00a0\\\\left( \\\\color{red}{\\\\underbrace{\\\\color{black}{\\\\frac{1}{\\\\alpha}\\\\texttt{e}^{\\\\alpha x}}}_{\\\\large v}} \\\\color{red}{\\\\underbrace{\\\\color{black}{\\\\texttt{cos}(\\\\beta x)}}_{\\\\large u}}\\\\:- \\\\int \\\\color{red}{\\\\underbrace{\\\\color{black}{\\\\left( \\\\frac{1}{\\\\alpha}\\\\texttt{e}^{\\\\alpha x} \\\\right)}}_{\\\\large v}} \\\\color{red}{\\\\underbrace{\\\\color{black}{(-\\\\beta \\\\texttt{sin}(\\\\beta x)dx)}}_{\\\\large du}}\\u00a0\\\\right)<br \\\/>&amp; \\\\color{gray}{\\\\text{[Apply the substitution]}} \\\\\\\\[5pt]<br \\\/>&amp;= \\\\frac{1}{\\\\alpha} \\\\texttt{e}^{\\\\alpha x}\\\\texttt{sin}(\\\\beta x)\\\\:- \\\\frac{\\\\beta}{\\\\alpha^2}\\\\texttt{e}^{\\\\alpha x}\\\\texttt{cos}(\\\\beta x)\\\\:- \\\\frac{\\\\beta^2}{\\\\alpha^2}\\\\int \\\\texttt{e}^{\\\\alpha x}\\\\texttt{sin}(\\\\beta x)dx<br \\\/>&amp; \\\\color{gray}{\\\\text{[Simplify]}}<br \\\/>\\\\end{align*}<br \\\/>\\\\)<p>\\u00a0<\\\/p><p><span style=\\\"font-size: 18.6667px;\\\">Given that we ended up with the same integral we wanted to solve for, we can equate the original integrand to the resulting expression and perform algebraic manipulations while treating the reappearing integral as a variable to solve for:<\\\/span><\\\/p>\\\\(\\\\normalsize<br \\\/>\\\\begin{align*}\\\\displaystyle<br \\\/>&amp; \\\\color{red}{\\\\int \\\\texttt{e}^{\\\\alpha x}\\\\texttt{sin}(\\\\beta x)dx} = \\\\frac{1}{\\\\alpha} \\\\texttt{e}^{\\\\alpha x}\\\\texttt{sin}(\\\\beta x)\\\\:- \\\\frac{\\\\beta}{\\\\alpha^2}\\\\texttt{e}^{\\\\alpha x}\\\\texttt{cos}(\\\\beta x)\\\\:- \\\\color{red}{\\\\frac{\\\\beta^2}{\\\\alpha^2}\\\\int \\\\texttt{e}^{\\\\alpha x}\\\\texttt{sin}(\\\\beta x)dx}<br \\\/>&amp; \\\\color{gray}{\\\\text{[Form an equation with both integrals]}} \\\\\\\\[5pt]<br \\\/>&amp; \\\\color{red}{\\\\frac{\\\\beta^2\\\\:+ \\\\alpha^2}{\\\\alpha^2} \\\\int \\\\texttt{e}^{\\\\alpha x}\\\\texttt{sin}(\\\\beta x)dx} = \\\\frac{1}{\\\\alpha} \\\\texttt{e}^{\\\\alpha x}\\\\texttt{sin}(\\\\beta x)\\\\:- \\\\frac{\\\\beta}{\\\\alpha^2}\\\\texttt{e}^{\\\\alpha x}\\\\texttt{cos}(\\\\beta x)\\\\:+ C<br \\\/>&amp; \\\\color{gray}{\\\\text{[Arrange both integrals to the left hand side of the equation]}} \\\\\\\\[5pt]<br \\\/>&amp; \\\\color{red}{\\\\int \\\\texttt{e}^{\\\\alpha x}\\\\texttt{sin}(\\\\beta x)dx} = \\\\left( \\\\frac{\\\\alpha^2}{\\\\alpha^2+\\\\beta^2} \\\\right) \\\\left( \\\\frac{1}{\\\\alpha}\\\\texttt{e}^{\\\\alpha x} \\\\right) [\\\\texttt{sin}(\\\\beta x)\\\\:- \\\\frac{\\\\beta}{\\\\alpha}\\\\texttt{cos}(\\\\beta x)]\\\\:+ C<br \\\/>&amp; \\\\color{gray}{\\\\text{[Factor]}} \\\\\\\\[5pt]<br \\\/>&amp; \\\\int \\\\texttt{e}^{\\\\alpha x}\\\\texttt{sin}(\\\\beta x)dx = \\\\left( \\\\frac{\\\\alpha}{\\\\alpha^2+\\\\beta^2} \\\\right) \\\\left(\\\\texttt{e}^{\\\\alpha x} \\\\right) [\\\\texttt{sin}(\\\\beta x)\\\\:- \\\\frac{\\\\beta}{\\\\alpha}\\\\texttt{cos}(\\\\beta x)]\\\\:+ C<br \\\/>&amp; \\\\color{gray}{\\\\text{[Simplify]}} \\\\\\\\[5pt]<br \\\/>&amp; \\\\color{green}{\\\\int \\\\texttt{e}^{\\\\alpha x}\\\\texttt{sin}(\\\\beta x)dx = \\\\left( \\\\frac{\\\\texttt{e}^{\\\\alpha x}}{\\\\alpha^2\\\\:+ \\\\beta^2} \\\\right)[\\\\alpha\\\\texttt{sin}(\\\\beta x)\\\\:- \\\\beta\\\\texttt{cos}(\\\\beta x)]\\\\:+ C}<br \\\/>&amp; \\\\color{gray}{\\\\text{[Distribute the numerator]}}<br \\\/>\\\\end{align*}<br \\\/>\\\\)\"}}]}<\/script>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t<section class=\"elementor-section elementor-top-section elementor-element elementor-element-5d255c34 elementor-section-boxed elementor-section-height-default elementor-section-height-default\" data-id=\"5d255c34\" data-element_type=\"section\">\n\t\t\t\t\t\t<div class=\"elementor-container elementor-column-gap-default\">\n\t\t\t\t\t<div class=\"elementor-column elementor-col-100 elementor-top-column elementor-element elementor-element-74f8b769\" data-id=\"74f8b769\" data-element_type=\"column\">\n\t\t\t<div class=\"elementor-widget-wrap elementor-element-populated\">\n\t\t\t\t\t\t<div class=\"elementor-element elementor-element-5af050df elementor-widget elementor-widget-toggle\" data-id=\"5af050df\" data-element_type=\"widget\" data-widget_type=\"toggle.default\">\n\t\t\t\t<div class=\"elementor-widget-container\">\n\t\t\t\t\t\t\t<div class=\"elementor-toggle\">\n\t\t\t\t\t\t\t<div class=\"elementor-toggle-item\">\n\t\t\t\t\t<h2 id=\"elementor-tab-title-1521\" class=\"elementor-tab-title\" data-tab=\"1\" role=\"button\" aria-controls=\"elementor-tab-content-1521\" aria-expanded=\"false\">\n\t\t\t\t\t\t\t\t\t\t\t\t<span class=\"elementor-toggle-icon elementor-toggle-icon-left\" aria-hidden=\"true\">\n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t<span class=\"elementor-toggle-icon-closed\"><i class=\"fas fa-toggle-off\"><\/i><\/span>\n\t\t\t\t\t\t\t\t<span class=\"elementor-toggle-icon-opened\"><i class=\"elementor-toggle-icon-opened fas fa-toggle-on\"><\/i><\/span>\n\t\t\t\t\t\t\t\t\t\t\t\t\t<\/span>\n\t\t\t\t\t\t\t\t\t\t\t\t<a class=\"elementor-toggle-title\" tabindex=\"0\">Problem 9: Solution<\/a>\n\t\t\t\t\t<\/h2>\n\n\t\t\t\t\t<div id=\"elementor-tab-content-1521\" class=\"elementor-tab-content elementor-clearfix\" data-tab=\"1\" role=\"region\" aria-labelledby=\"elementor-tab-title-1521\"><p><span style=\"font-size: 18.6667px;\">Let \\(u=\\texttt{cos}(\\beta x)\\), so \\(du=-\\beta \\texttt{sin}(\\beta x)dx\\); and \\(dv=\\texttt{e}^{\\alpha x}dx\\), so \\(v=\\frac{1}{\\alpha}\\texttt{e}^{\\alpha x}\\).<\/span><span style=\"font-size: 18.6667px;\">&nbsp;<\/span><span style=\"font-size: 18.6667px;\">Substituting these values into \\(uv\\:- \\int vdu\\) leads to the following:<\/span><\/p>\n\\(\\normalsize<br>\\begin{align*}\\displaystyle<br>\\int \\texttt{e}^{\\alpha x}\\texttt{cos}(\\beta x)dx<br>&amp;= \\color{red}{\\underbrace{\\color{black}{\\frac{1}{\\alpha}\\texttt{e}^{\\alpha x}}}_{\\large v}} \\color{red}{\\underbrace{\\color{black}{\\texttt{cos}(\\beta x)}}_{\\large u}}\\:- \\int \\color{red}{\\underbrace{\\color{black}{\\left( \\frac{1}{\\alpha}\\texttt{e}^{\\alpha x} \\right)}}_{\\large v}} \\color{red}{\\underbrace{\\color{black}{(-\\beta \\texttt{sin}(\\beta x)dx)}}_{\\large du}}<br>&amp; \\color{gray}{\\text{[Apply the substitution]}} \\\\[5pt]<br>&amp;= \\frac{1}{\\alpha}\\texttt{e}^{\\alpha x}\\texttt{cos}(\\beta x)\\:+ \\frac{\\beta}{\\alpha}\\int \\texttt{e}^{\\alpha x}\\texttt{sin}(\\beta x)dx<br>&amp; \\color{gray}{\\text{[Simplify]}} \\\\[5pt]<br>\\end{align*}<br>\\)\n<p><\/p>\n<p><span style=\"font-size: 18.6667px;\">Use IBP again.&nbsp;<\/span><span style=\"font-size: 18.6667px;\">Let \\(u=\\texttt{sin}(\\beta x)\\), so \\(du=\\beta \\texttt{cos}(\\beta x)dx\\); and \\(dv=\\texttt{e}^{\\alpha x}dx\\), so \\(v=\\frac{1}{\\alpha }\\texttt{e}^{\\alpha x}\\).<\/span><span style=\"font-size: 18.6667px;\">&nbsp;<\/span><span style=\"font-size: 18.6667px;\">Substituting these values into \\(uv\\:- \\int vdu\\) leads to the following:<\/span><\/p>\n\\(\\normalsize<br>\\begin{align*}\\displaystyle<br>\\frac{1}{\\alpha}\\texttt{e}^{\\alpha x}\\texttt{cos}(\\beta x)\\:+ \\frac{\\beta}{\\alpha}\\int \\texttt{e}^{\\alpha x}\\texttt{sin}(\\beta x)dx<br>&amp;=\\frac{1}{\\alpha}\\texttt{e}^{\\alpha x}\\texttt{cos}(\\beta x)\\:+ \\frac{\\beta}{\\alpha}&nbsp;\\left( \\color{red}{\\underbrace{\\color{black}{\\frac{1}{\\alpha}\\texttt{e}^{\\alpha x}}}_{\\large v}} \\color{red}{\\underbrace{\\color{black}{\\texttt{sin}(\\beta x)}}_{\\large u}}\\:- \\int \\color{red}{\\underbrace{\\color{black}{\\left( \\frac{1}{\\alpha}\\texttt{e}^{\\alpha x} \\right)}}_{\\large v}} \\color{red}{\\underbrace{\\color{black}{(\\beta \\texttt{cos}(\\beta x)dx)}}_{\\large du}}&nbsp;\\right)<br>&amp; \\color{gray}{\\text{[Apply the substitution]}} \\\\[5pt]<br>&amp;= \\frac{1}{\\alpha} \\texttt{e}^{\\alpha x}\\texttt{cos}(\\beta x)\\:+ \\frac{\\beta}{\\alpha^2}\\texttt{e}^{\\alpha x}\\texttt{sin}(\\beta x)\\:- \\frac{\\beta^2}{\\alpha^2}\\int \\texttt{e}^{\\alpha x}\\texttt{cos}(\\beta x)dx<br>&amp; \\color{gray}{\\text{[Simplify]}}<br>\\end{align*}<br>\\)\n<p><\/p>\n<p><span style=\"font-size: 18.6667px;\">Given that we ended up with the same integral we wanted to solve for, we can equate the original integrand to the resulting expression and perform algebraic manipulations while treating the reappearing integral as a variable to solve for:<\/span><\/p>\n\\(\\normalsize<br>\\begin{align*}\\displaystyle<br>&amp; \\color{red}{\\int \\texttt{e}^{\\alpha x}\\texttt{cos}(\\beta x)dx} = \\frac{1}{\\alpha} \\texttt{e}^{\\alpha x}\\texttt{cos}(\\beta x)\\:+ \\frac{\\beta}{\\alpha^2}\\texttt{e}^{\\alpha x}\\texttt{sin}(\\beta x)\\:- \\color{red}{\\frac{\\beta^2}{\\alpha^2}\\int \\texttt{e}^{\\alpha x}\\texttt{cos}(\\beta x)dx}<br>&amp; \\color{gray}{\\text{[Form an equation with both integrals]}} \\\\[5pt]<br>&amp; \\color{red}{\\frac{\\beta^2\\:+ \\alpha^2}{\\alpha^2} \\int \\texttt{e}^{\\alpha x}\\texttt{cos}(\\beta x)dx} = \\frac{1}{\\alpha} \\texttt{e}^{\\alpha x}\\texttt{cos}(\\beta x)\\:+ \\frac{\\beta}{\\alpha^2}\\texttt{e}^{\\alpha x}\\texttt{sin}(\\beta x)\\:+ C<br>&amp; \\color{gray}{\\text{[Arrange both integrals to the left hand side of the equation]}} \\\\[5pt]<br>&amp; \\color{red}{\\int \\texttt{e}^{\\alpha x}\\texttt{cos}(\\beta x)dx} = \\left( \\frac{\\alpha^2}{\\alpha^2+\\beta^2} \\right) \\left( \\frac{1}{\\alpha}\\texttt{e}^{\\alpha x} \\right) [\\texttt{cos}(\\beta x)\\:+ \\frac{\\beta}{\\alpha}\\texttt{sin}(\\beta x)]\\:+ C<br>&amp; \\color{gray}{\\text{[Factor]}} \\\\[5pt]<br>&amp; \\int \\texttt{e}^{\\alpha x}\\texttt{cos}(\\beta x)dx = \\left( \\frac{\\alpha}{\\alpha^2+\\beta^2} \\right) \\left(\\texttt{e}^{\\alpha x} \\right) [\\texttt{cos}(\\beta x)\\:+ \\frac{\\beta}{\\alpha}\\texttt{sin}(\\beta x)]\\:+ C<br>&amp; \\color{gray}{\\text{[Simplify]}} \\\\[5pt]<br>&amp; \\color{green}{\\int \\texttt{e}^{\\alpha x}\\texttt{cos}(\\beta x)dx = \\left( \\frac{\\texttt{e}^{\\alpha x}}{\\alpha^2\\:+ \\beta^2} \\right)[\\alpha\\texttt{cos}(\\beta x)\\:+ \\beta\\texttt{sin}(\\beta x)]\\:+ C}<br>&amp; \\color{gray}{\\text{[Distribute the numerator]}}<br>\\end{align*}<br>\\)<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t\t\t\t\t<script type=\"application\/ld+json\">{\"@context\":\"https:\\\/\\\/schema.org\",\"@type\":\"FAQPage\",\"mainEntity\":[{\"@type\":\"Question\",\"name\":\"Problem 9: Solution\",\"acceptedAnswer\":{\"@type\":\"Answer\",\"text\":\"<p><span style=\\\"font-size: 18.6667px;\\\">Let \\\\(u=\\\\texttt{cos}(\\\\beta x)\\\\), so \\\\(du=-\\\\beta \\\\texttt{sin}(\\\\beta x)dx\\\\); and \\\\(dv=\\\\texttt{e}^{\\\\alpha x}dx\\\\), so \\\\(v=\\\\frac{1}{\\\\alpha}\\\\texttt{e}^{\\\\alpha x}\\\\).<\\\/span><span style=\\\"font-size: 18.6667px;\\\">&nbsp;<\\\/span><span style=\\\"font-size: 18.6667px;\\\">Substituting these values into \\\\(uv\\\\:- \\\\int vdu\\\\) leads to the following:<\\\/span><\\\/p>\\n\\\\(\\\\normalsize<br>\\\\begin{align*}\\\\displaystyle<br>\\\\int \\\\texttt{e}^{\\\\alpha x}\\\\texttt{cos}(\\\\beta x)dx<br>&amp;= \\\\color{red}{\\\\underbrace{\\\\color{black}{\\\\frac{1}{\\\\alpha}\\\\texttt{e}^{\\\\alpha x}}}_{\\\\large v}} \\\\color{red}{\\\\underbrace{\\\\color{black}{\\\\texttt{cos}(\\\\beta x)}}_{\\\\large u}}\\\\:- \\\\int \\\\color{red}{\\\\underbrace{\\\\color{black}{\\\\left( \\\\frac{1}{\\\\alpha}\\\\texttt{e}^{\\\\alpha x} \\\\right)}}_{\\\\large v}} \\\\color{red}{\\\\underbrace{\\\\color{black}{(-\\\\beta \\\\texttt{sin}(\\\\beta x)dx)}}_{\\\\large du}}<br>&amp; \\\\color{gray}{\\\\text{[Apply the substitution]}} \\\\\\\\[5pt]<br>&amp;= \\\\frac{1}{\\\\alpha}\\\\texttt{e}^{\\\\alpha x}\\\\texttt{cos}(\\\\beta x)\\\\:+ \\\\frac{\\\\beta}{\\\\alpha}\\\\int \\\\texttt{e}^{\\\\alpha x}\\\\texttt{sin}(\\\\beta x)dx<br>&amp; \\\\color{gray}{\\\\text{[Simplify]}} \\\\\\\\[5pt]<br>\\\\end{align*}<br>\\\\)\\n<p><\\\/p>\\n<p><span style=\\\"font-size: 18.6667px;\\\">Use IBP again.&nbsp;<\\\/span><span style=\\\"font-size: 18.6667px;\\\">Let \\\\(u=\\\\texttt{sin}(\\\\beta x)\\\\), so \\\\(du=\\\\beta \\\\texttt{cos}(\\\\beta x)dx\\\\); and \\\\(dv=\\\\texttt{e}^{\\\\alpha x}dx\\\\), so \\\\(v=\\\\frac{1}{\\\\alpha }\\\\texttt{e}^{\\\\alpha x}\\\\).<\\\/span><span style=\\\"font-size: 18.6667px;\\\">&nbsp;<\\\/span><span style=\\\"font-size: 18.6667px;\\\">Substituting these values into \\\\(uv\\\\:- \\\\int vdu\\\\) leads to the following:<\\\/span><\\\/p>\\n\\\\(\\\\normalsize<br>\\\\begin{align*}\\\\displaystyle<br>\\\\frac{1}{\\\\alpha}\\\\texttt{e}^{\\\\alpha x}\\\\texttt{cos}(\\\\beta x)\\\\:+ \\\\frac{\\\\beta}{\\\\alpha}\\\\int \\\\texttt{e}^{\\\\alpha x}\\\\texttt{sin}(\\\\beta x)dx<br>&amp;=\\\\frac{1}{\\\\alpha}\\\\texttt{e}^{\\\\alpha x}\\\\texttt{cos}(\\\\beta x)\\\\:+ \\\\frac{\\\\beta}{\\\\alpha}&nbsp;\\\\left( \\\\color{red}{\\\\underbrace{\\\\color{black}{\\\\frac{1}{\\\\alpha}\\\\texttt{e}^{\\\\alpha x}}}_{\\\\large v}} \\\\color{red}{\\\\underbrace{\\\\color{black}{\\\\texttt{sin}(\\\\beta x)}}_{\\\\large u}}\\\\:- \\\\int \\\\color{red}{\\\\underbrace{\\\\color{black}{\\\\left( \\\\frac{1}{\\\\alpha}\\\\texttt{e}^{\\\\alpha x} \\\\right)}}_{\\\\large v}} \\\\color{red}{\\\\underbrace{\\\\color{black}{(\\\\beta \\\\texttt{cos}(\\\\beta x)dx)}}_{\\\\large du}}&nbsp;\\\\right)<br>&amp; \\\\color{gray}{\\\\text{[Apply the substitution]}} \\\\\\\\[5pt]<br>&amp;= \\\\frac{1}{\\\\alpha} \\\\texttt{e}^{\\\\alpha x}\\\\texttt{cos}(\\\\beta x)\\\\:+ \\\\frac{\\\\beta}{\\\\alpha^2}\\\\texttt{e}^{\\\\alpha x}\\\\texttt{sin}(\\\\beta x)\\\\:- \\\\frac{\\\\beta^2}{\\\\alpha^2}\\\\int \\\\texttt{e}^{\\\\alpha x}\\\\texttt{cos}(\\\\beta x)dx<br>&amp; \\\\color{gray}{\\\\text{[Simplify]}}<br>\\\\end{align*}<br>\\\\)\\n<p><\\\/p>\\n<p><span style=\\\"font-size: 18.6667px;\\\">Given that we ended up with the same integral we wanted to solve for, we can equate the original integrand to the resulting expression and perform algebraic manipulations while treating the reappearing integral as a variable to solve for:<\\\/span><\\\/p>\\n\\\\(\\\\normalsize<br>\\\\begin{align*}\\\\displaystyle<br>&amp; \\\\color{red}{\\\\int \\\\texttt{e}^{\\\\alpha x}\\\\texttt{cos}(\\\\beta x)dx} = \\\\frac{1}{\\\\alpha} \\\\texttt{e}^{\\\\alpha x}\\\\texttt{cos}(\\\\beta x)\\\\:+ \\\\frac{\\\\beta}{\\\\alpha^2}\\\\texttt{e}^{\\\\alpha x}\\\\texttt{sin}(\\\\beta x)\\\\:- \\\\color{red}{\\\\frac{\\\\beta^2}{\\\\alpha^2}\\\\int \\\\texttt{e}^{\\\\alpha x}\\\\texttt{cos}(\\\\beta x)dx}<br>&amp; \\\\color{gray}{\\\\text{[Form an equation with both integrals]}} \\\\\\\\[5pt]<br>&amp; \\\\color{red}{\\\\frac{\\\\beta^2\\\\:+ \\\\alpha^2}{\\\\alpha^2} \\\\int \\\\texttt{e}^{\\\\alpha x}\\\\texttt{cos}(\\\\beta x)dx} = \\\\frac{1}{\\\\alpha} \\\\texttt{e}^{\\\\alpha x}\\\\texttt{cos}(\\\\beta x)\\\\:+ \\\\frac{\\\\beta}{\\\\alpha^2}\\\\texttt{e}^{\\\\alpha x}\\\\texttt{sin}(\\\\beta x)\\\\:+ C<br>&amp; \\\\color{gray}{\\\\text{[Arrange both integrals to the left hand side of the equation]}} \\\\\\\\[5pt]<br>&amp; \\\\color{red}{\\\\int \\\\texttt{e}^{\\\\alpha x}\\\\texttt{cos}(\\\\beta x)dx} = \\\\left( \\\\frac{\\\\alpha^2}{\\\\alpha^2+\\\\beta^2} \\\\right) \\\\left( \\\\frac{1}{\\\\alpha}\\\\texttt{e}^{\\\\alpha x} \\\\right) [\\\\texttt{cos}(\\\\beta x)\\\\:+ \\\\frac{\\\\beta}{\\\\alpha}\\\\texttt{sin}(\\\\beta x)]\\\\:+ C<br>&amp; \\\\color{gray}{\\\\text{[Factor]}} \\\\\\\\[5pt]<br>&amp; \\\\int \\\\texttt{e}^{\\\\alpha x}\\\\texttt{cos}(\\\\beta x)dx = \\\\left( \\\\frac{\\\\alpha}{\\\\alpha^2+\\\\beta^2} \\\\right) \\\\left(\\\\texttt{e}^{\\\\alpha x} \\\\right) [\\\\texttt{cos}(\\\\beta x)\\\\:+ \\\\frac{\\\\beta}{\\\\alpha}\\\\texttt{sin}(\\\\beta x)]\\\\:+ C<br>&amp; \\\\color{gray}{\\\\text{[Simplify]}} \\\\\\\\[5pt]<br>&amp; \\\\color{green}{\\\\int \\\\texttt{e}^{\\\\alpha x}\\\\texttt{cos}(\\\\beta x)dx = \\\\left( \\\\frac{\\\\texttt{e}^{\\\\alpha x}}{\\\\alpha^2\\\\:+ \\\\beta^2} \\\\right)[\\\\alpha\\\\texttt{cos}(\\\\beta x)\\\\:+ \\\\beta\\\\texttt{sin}(\\\\beta x)]\\\\:+ C}<br>&amp; \\\\color{gray}{\\\\text{[Distribute the numerator]}}<br>\\\\end{align*}<br>\\\\)\"}}]}<\/script>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t<section class=\"elementor-section elementor-top-section elementor-element elementor-element-53bf611 elementor-section-boxed elementor-section-height-default elementor-section-height-default\" data-id=\"53bf611\" data-element_type=\"section\">\n\t\t\t\t\t\t<div class=\"elementor-container elementor-column-gap-default\">\n\t\t\t\t\t<div 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menu-item-12217\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/limits\/lhopitals-rule\/\" class=\"elementor-sub-item\" tabindex=\"-1\">L\u2019H\u00f4pital\u2019s Rule<\/a><\/li>\n\t<\/ul>\n<\/li>\n\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-12629\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/limits\/virtual-lessons-limits\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Virtual Lessons<\/a><\/li>\n<\/ul>\n<\/li>\n<li class=\"menu-item menu-item-type-custom menu-item-object-custom menu-item-has-children menu-item-14327\"><a href=\"#\" class=\"elementor-item elementor-item-anchor\" tabindex=\"-1\">Differentiation<\/a>\n<ul class=\"sub-menu elementor-nav-menu--dropdown\">\n\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-14856\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/differentiation\/fundamentals-differentiation\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Fundamentals<\/a><\/li>\n\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-11537\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/differentiation\/differentiation-formulas-update\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Differentiation Formulas<\/a><\/li>\n\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-11409\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/differentiation\/differentiation-techniques\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Differentiation Techniques<\/a><\/li>\n\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-has-children menu-item-11411\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/differentiation\/differentiation-applications\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Differentiation Applications<\/a>\n\t<ul 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menu-item-type-post_type menu-item-object-page menu-item-20868\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/differentiation\/problem-set-differentiation\/\" class=\"elementor-sub-item\" tabindex=\"-1\"><img decoding=\"async\" src=\"https:\/\/onlinemathtutor.co\/wp-content\/uploads\/2022\/02\/Problem-Sets.png\"  width=\"20\" style=\"margin-right: 10px;\">  Problem Set<\/a><\/li>\n<\/ul>\n<\/li>\n<li class=\"menu-item menu-item-type-custom menu-item-object-custom menu-item-has-children menu-item-14329\"><a href=\"#\" class=\"elementor-item elementor-item-anchor\" tabindex=\"-1\">Integration<\/a>\n<ul class=\"sub-menu elementor-nav-menu--dropdown\">\n\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-has-children menu-item-14551\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/fundamentals-integration\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Fundamentals<\/a>\n\t<ul class=\"sub-menu elementor-nav-menu--dropdown\">\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-10226\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/riemann-sums\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Riemann Sums<\/a><\/li>\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-11365\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/fundamental-theorem-of-calculus\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Fundamental Theorem of Calculus<\/a><\/li>\n\t<\/ul>\n<\/li>\n\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-11530\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/integration-formulas-update\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Integration Formulas<\/a><\/li>\n\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-9815\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/integration-techinques\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Integration Techinques<\/a><\/li>\n\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-13700\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/integration-applications\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Integration Applications<\/a><\/li>\n\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-19550\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/\" class=\"elementor-sub-item\" tabindex=\"-1\"><img decoding=\"async\" src=\"https:\/\/onlinemathtutor.co\/wp-content\/uploads\/2022\/02\/Problem-Sets.png\"  width=\"20\" style=\"margin-right: 10px;\">  Problem Set<\/a><\/li>\n<\/ul>\n<\/li>\n<li class=\"menu-item menu-item-type-custom menu-item-object-custom menu-item-has-children menu-item-14330\"><a href=\"#\" class=\"elementor-item elementor-item-anchor\" tabindex=\"-1\">Sequences and Series<\/a>\n<ul class=\"sub-menu elementor-nav-menu--dropdown\">\n\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-11364\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/sequences\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Fundamentals<\/a><\/li>\n\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-9804\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/sequences\/convergence-test\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Convergence Test<\/a><\/li>\n\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-10233\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/sequences\/power-series\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Power Series<\/a><\/li>\n\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-11363\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/sequences\/taylor-series\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Taylor Series<\/a><\/li>\n\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-21607\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/sequences\/problem-set-sequences-series\/\" class=\"elementor-sub-item\" tabindex=\"-1\"><img decoding=\"async\" src=\"https:\/\/onlinemathtutor.co\/wp-content\/uploads\/2022\/02\/Problem-Sets.png\"  width=\"20\" style=\"margin-right: 10px;\">  Problem Set<\/a><\/li>\n<\/ul>\n<\/li>\n<li class=\"menu-item menu-item-type-custom menu-item-object-custom menu-item-has-children menu-item-14331\"><a href=\"#\" class=\"elementor-item elementor-item-anchor\" tabindex=\"-1\">Parametric Equations<\/a>\n<ul class=\"sub-menu elementor-nav-menu--dropdown\">\n\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-12174\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/parametric-equations-2\/fundamentals-of-parametric-equations\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Fundamentals<\/a><\/li>\n\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-19376\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/parametric-equations-2\/polar-coordinates-calculus\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Polar Coordinates<\/a><\/li>\n\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-11366\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/precalculus-resources\/unit-circle\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Unit Circle<\/a><\/li>\n\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-12198\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/parametric-equations-2\/calculus-parametric-equation\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Calculus of Parametric Equations<\/a><\/li>\n\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-has-children menu-item-21606\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/parametric-equations-2\/problem-set-parametric-eqs-2\/\" class=\"elementor-sub-item\" tabindex=\"-1\"><img decoding=\"async\" src=\"https:\/\/onlinemathtutor.co\/wp-content\/uploads\/2022\/02\/Problem-Sets.png\"  width=\"20\" style=\"margin-right: 10px;\">  Problem Set<\/a>\n\t<ul class=\"sub-menu elementor-nav-menu--dropdown\">\n\t\t<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-12615\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/parametric-equations-2\/virtual-lessons-parametric-equations\/\" class=\"elementor-sub-item\" tabindex=\"-1\">Virtual Lessons<\/a><\/li>\n\t<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n<li class=\"menu-item menu-item-type-post_type menu-item-object-page menu-item-10232\"><a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/ap-calculus-frqs\/\" class=\"elementor-item\" tabindex=\"-1\">AP Calculus Prep<\/a><\/li>\n<\/ul>\t\t\t<\/nav>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t<div class=\"elementor-column elementor-col-50 elementor-top-column elementor-element elementor-element-6766c807\" data-id=\"6766c807\" data-element_type=\"column\">\n\t\t\t<div class=\"elementor-widget-wrap elementor-element-populated\">\n\t\t\t\t\t\t<div class=\"elementor-element elementor-element-37924575 elementor-widget elementor-widget-spacer\" data-id=\"37924575\" data-element_type=\"widget\" data-widget_type=\"spacer.default\">\n\t\t\t\t<div class=\"elementor-widget-container\">\n\t\t\t\t\t\t\t<div class=\"elementor-spacer\">\n\t\t\t<div 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This technique is useful whenever a product appears in the integrand and the implement u-substitution is not feasible. Virtual Lessons <a href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/integration-practice-problems-ibp\/\" class=\"read-more\">Read More &#8230;<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"parent":19541,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"elementor_header_footer","meta":{"footnotes":""},"class_list":["post-19689","page","type-page","status-publish","hentry"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v24.6 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Integration Practice Problems - [IBP] | Online Math Tutoring LLC | Expert Math Tutor Near Me<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/onlinemathtutor.co\/course-content\/calculus-resources\/integration\/problem-set-integration\/integration-practice-problems-ibp\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Integration Practice Problems - [IBP] | Online Math Tutoring LLC | Expert Math Tutor Near Me\" \/>\n<meta property=\"og:description\" content=\"Practice Problems: Integation By Parts Integration by parts is an integration technique that involves separating the integrand into two parts. 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