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Practice Problems: Trigonometric Integrals

When integrating products of trigonometric functions, the general practice involves applying the trigonometric versions of the Pythagorean Theorem such as 

\(\texttt{sin}^2(\theta)+\texttt{cos}^2(\theta)=1\)    or    \(\texttt{tan}^2(\theta)+1=\texttt{sec}^2(\theta)\)

in conjunction with an appropriate u-substitution. If the powers both even then you may need to consider applying a Power Reduction Identitiy (Half-Angle Idenitty). 

\(\texttt{sin}(2\theta)\frac{1-\texttt{cos}(2\theta)}{2}\)    or  \(\texttt{cos}(2\theta)\frac{1+\texttt{cos}(2\theta)}{2}\)

Virtual Lessons

Need some additional help understanding how to apply this integration technique? Click Here to visit the virtual lesson section.

Practice Problems

Answer the following questions by employing the Pythagorean Theorem or the Power Reduction identities. 

  1. \(\displaystyle \int \texttt{sin}^2(x)\texttt{cos}^4(x) dx \)

  2. \(\displaystyle \int \texttt{sin}^3(x)\texttt{cos}^4(x) dx \)

  3. \(\displaystyle \int \texttt{sin}^3(x) dx \)

  4. \(\displaystyle \int^{\pi}_{-\pi} \texttt{sin}(3x)\texttt{cos}(2x) dx \)
  1. \(\displaystyle \int^{\frac{\pi}{4}}_{0}  \texttt{sin}^3(x) dx \)

  2. \(\displaystyle \int \texttt{sec}(4x) dx \)

  3. \(\displaystyle \int \texttt{tan}^3(3x)\texttt{sec}^2(3x) dx \)

  4. \(\displaystyle \int \texttt{tan}^4(x)\texttt{sec}^2(x) dx \)
  1. \(\displaystyle \int \frac{1+\texttt{cos}(2x)}{1-\texttt{cos}(2x)} dx \)

  2. \(\displaystyle \int \sqrt{1-\texttt{cos}(\theta)} d\theta \)

Suggestive Solution Guide

Use the double angle identity for sines: \(2\texttt{sin}(x)\texttt{cos}(x) = \texttt{sin}(2x)\), so \(\texttt{sin}(x)\texttt{cos}(x)=\frac{1}{2}\texttt{sin}(2x)\):

 

\(\normalsize
\begin{align*}\displaystyle
\int \texttt{sin}^2(x)\texttt{cos}^4(x) dx
&= \int [\color{red}{\underbrace{\color{black}{\texttt{sin}(x)\texttt{cos}(x)}}_{\large \frac{1}{2}\texttt{sin}(2x)}}]^2 \texttt{cos}^2(x)dx
& \color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= \frac{1}{4}\int \texttt{sin}^2(2x)\texttt{cos}^2(x)dx
& \color{gray}{\text{[Rearrange the expression]}} \\[5pt]
\end{align*}
\)

 

 

Apply the power reduction identity for cosines: \(\texttt{cos}^2(x) = \frac{1\:+ \texttt{cos}(2x)}{2}\). Substituting yields:

 

\(\normalsize
\begin{align*}\displaystyle
\frac{1}{4}\int \texttt{sin}^2(2x)\texttt{cos}^2(x)dx
&= \frac{1}{4}\int [\texttt{sin}^2(2x)] \left[ \color{red}{\underbrace{\color{black}{\frac{1\:+ \texttt{cos}(2x)}{2}}}_{\texttt{cos}^2(x)}}dx \right]
& \color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= \frac{1}{8}\left( \int \texttt{sin}^2(2x)dx\:+ \int \texttt{sin}^2(2x)\texttt{cos}(2x)dx \right)
& \color{gray}{\text{[Rearrange the expression]}} \\[5pt]
\end{align*}
\)

 

 

Use a quick U-sub. Let \(u=2x\), so \(du=2dx\). Substituting yields:

\(\normalsize
\begin{align*}\displaystyle
\frac{1}{4}\int \texttt{sin}^2(2x)\texttt{cos}^2(x)dx
&= \frac{1}{8}\left( \int \texttt{sin}^2(\color{red}{\underbrace{\color{black}{2x}}_{\large u}}) \color{red}{\underbrace{\color{black}{dx}}_{\large \frac{du}{2}}} \:+ \int \texttt{sin}^2(\color{red}{\underbrace{\color{black}{2x}}_{\large u}})\texttt{cos}(\color{red}{\underbrace{\color{black}{2x}}_{\large u}}) \color{red}{\underbrace{\color{black}{dx}}_{\large \frac{du}{2}}} \right)
& \color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= \frac{1}{16}\left( \int \texttt{sin}^2(u)du\:+ \int \texttt{sin}^2(u)\texttt{cos}(u)du \right)
& \color{gray}{\text{[Factor and rearrange the expression]}} \\[5pt]
\end{align*}
\)

 

 

Apply the power reduction identity for sines in the first integrand: \(\texttt{sin}^2(x) = \frac{1\:- \texttt{cos}(2x)}{2}\). Substituting yields:

\(\normalsize
\begin{align*}\displaystyle
\frac{1}{16}\left( \int \texttt{sin}^2(u)du\:+ \int \texttt{sin}^2(u)\texttt{cos}(u)du \right)
&= \frac{1}{16}\left( \int \color{red}{\underbrace{\color{black}{\texttt{sin}^2(u)}}_{\large \frac{1-\texttt{cos}(2u)}{2}}} \:+ \int \texttt{sin}^2(u)\texttt{cos}(u)du \right)
& \color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= \frac{1}{16}\left( \frac{1}{2}\int du\:- \frac{1}{2}\int \texttt{cos}(2u)du \:+ \int \texttt{sin}^2(u)\texttt{cos}(u)du \right)
& \color{gray}{\text{[Distribute the first integral]}} \\[5pt]
&= \frac{1}{32}u\:- \frac{1}{64}\texttt{sin}(2u)\:+ \frac{1}{16}\int \texttt{sin}^2(u)\texttt{cos}(u)du
& \color{gray}{\text{[Integrate (direct sub and u-sub)]}} \\[5pt]
&= \frac{1}{32}(2x)\:- \frac{1}{64}\texttt{sin}(2 \cdot 2x)\:+ \frac{1}{16}\int \texttt{sin}^2(u)\texttt{cos}(u)du
& \color{gray}{\text{[Substitute back to x]}} \\[5pt]
&= \frac{x}{16}\:- \frac{1}{64}\texttt{sin}(4x)\:+ \frac{1}{16} \color{red}{\int \texttt{sin}^2(u)\texttt{cos}(u)du}
& \color{gray}{\text{[Simplify]}} \\[5pt]
\end{align*}
\)

 

 

Use a U-sub for the third integral. Let \(v=\texttt{sin}(u)\), so \(dv=\texttt{cos}(u)du\):

\(\normalsize
\begin{align*}\displaystyle
\color{red}{\int \texttt{sin}^2(u)\texttt{cos}(u)du}
&= \int \color{red}{\underbrace{\color{black}{\texttt{sin}^2(u)}}_{\large v^2}} \color{red}{\underbrace{\color{black}{\texttt{cos}(u)du}}_{\large dv}} 
& \color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= \int v^2dv = \frac{1}{3} v^3\:+ C
& \color{gray}{\text{[Integrate]}} \\[5pt]
&= \frac{1}{3} \texttt{sin}^3(u)\:+ C
& \color{gray}{\text{[Substitute back to u]}} \\[5pt]
&= \color{red}{\frac{1}{3} \texttt{sin}^3(2x)\:+ C}
& \color{gray}{\text{[Substitute back to x]}} \\[5pt]
\end{align*}
\)

 

 

Combine previous results into one expression:

\(\normalsize
\begin{align*}\displaystyle
\int \texttt{sin}^2(x)\texttt{cos}^4(x) dx
&= \frac{x}{16}\:- \frac{1}{64}\texttt{sin}(4x)\:+ \frac{1}{16} \color{red}{\int \texttt{sin}^2(u)\texttt{cos}(u)du}
& \color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= \frac{x}{16}\:- \frac{1}{64}\texttt{sin}(4x)\:+ \frac{1}{16} \left(\color{red}{\frac{1}{3} \texttt{sin}^3(2x)}\right)\:+ C
& \color{gray}{\text{[Rearrange the expression]}} \\[5pt]
&= \color{green}{\frac{1}{48}\texttt{sin}^3(2x)\:- \frac{1}{64}\texttt{sin}(4x)\:+ \frac{x}{16}\:+ C}
\end{align*}
\)

Apply the pythagoras theorem: \(\texttt{sin}^2(x)+\texttt{cos}^2(x) = 1\), so \(\texttt{sin}^2(x) = 1\:- \texttt{cos}^2(x)\)

 

\(\normalsize
\begin{align*}\displaystyle
\int \texttt{sin}^3(x)\texttt{cos}^4(x) dx = \int \texttt{sin}(x)[1\:- \texttt{cos}^2(x)]\texttt{cos}^4(x)dx
\end{align*}
\)

 

 

Apply a U-Sub. Let \(u=\texttt{cos}(x)\), so \(du=-\texttt{sin}(x)dx\). Substituting these values leads to:

\(\normalsize
\begin{align*}\displaystyle
\int \texttt{sin}(x)[1\:- \texttt{cos}^2(x)]\texttt{cos}^4(x)dx
&= \int \color{red}{\underbrace{\color{black}{\texttt{cos}^4}}_{\large u^4}} \left[ 1-\color{red}{\underbrace{\color{black}{\texttt{cos}^2(x)dx}}_{\large u^2}} \right] \color{red}{\underbrace{\color{black}{\texttt{sin}(x)dx}}_{\large -du}}
& \color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= -\int u^4(1-u^2)du
& \color{gray}{\text{[Rearrange the expression]}} \\[5pt]
&= -\int u^4du\:+ \int u^6du
& \color{gray}{\text{[Separate the integrals]}} \\[5pt]
&= -\frac{1}{5}u^5\:+ \frac{1}{7}u^7\:+ C
& \color{gray}{\text{[Integrate]}} \\[5pt]
&= \color{green}{\frac{1}{7}\texttt{cos}^7(x)\:- \frac{1}{5}\texttt{cos}^5(x)\:+ C}
& \color{gray}{\text{[Substitute back to x and rearrange]}} \\[5pt]
\end{align*}
\)

 

Apply the pythagoras theorem: \(\texttt{sin}^2(x)+\texttt{cos}^2(x) = 1\), so \(\texttt{sin}^2(x) = 1\:- \texttt{cos}^2(x)\)

 

\(\normalsize
\begin{align*}\displaystyle
\int \texttt{sin}^3(x)dx = \int \texttt{sin}(x) \cdot \texttt{sin}^2(x) = \int \texttt{sin}(x)[1\:- \texttt{cos}^2(x)]dx
\end{align*}
\)

 

 

Apply a U-Sub. Let \(u=\texttt{cos}(x)\), so \(du=-\texttt{sin}(x)dx\). Substituting these values leads to:

\(\normalsize
\begin{align*}\displaystyle
\int [1-\texttt{cos}^2(x)]\texttt{sin}(x)dx
&= \int [1-\color{red}{\underbrace{\color{black}{\texttt{cos}^2(x)}}_{\large u^2}}] \color{red}{\underbrace{\color{black}{\texttt{sin}(x)}}_{\large -du}}
& \color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= -\int (1-u^2)du
& \color{gray}{\text{[Rearrange the expression]}} \\[5pt]
&= -\int du\:+ \int u^2du
& \color{gray}{\text{[Separate the integrals]}} \\[5pt]
&= -u\:+ \frac{1}{3}u^3\:+ C
& \color{gray}{\text{[Integrate]}} \\[5pt]
&= \color{green}{\frac{1}{3}\texttt{cos}^3(x)\:- \texttt{cos}(x)\:+ C}
& \color{gray}{\text{[Substitute back to x and rearrange]}} \\[5pt]
\end{align*}
\)

Apply the sine and cosine product identity: \(\texttt{sin}(a)\texttt{cos}(b) = \frac{1}{2} \left( \texttt{sin}(a+b) + \texttt{sin}(a-b) \right)\)

 

\(\normalsize
\begin{align*}\displaystyle
\int^{\pi}_{-\pi} \texttt{sin}(3x)\texttt{cos}(2x)dx
&= \frac{1}{2}\int^{\pi}_{-\pi} [\texttt{sin}(5x) + \texttt{sin}(x)]dx
& \color{gray}{\text{[Apply the identity]}} \\[5pt]
&= \frac{1}{2} \left(\int^{\pi}_{-\pi} \texttt{sin}(5x)dx\:+ \int^{\pi}_{-\pi} \texttt{sin}(x)dx \right)
& \color{gray}{\text{[Separate the integrand]}} \\[5pt]
&= \frac{1}{2} \left[ -\frac{1}{5}\texttt{cos}(5x)\:- \texttt{cos}(x) \right] \Bigr|^{\pi}_{-\pi}
& \color{gray}{\text{[Integrate using a quick U-sub]}} \\[5pt]
&= \frac{1}{2} \left( \left[ -\frac{1}{5}\texttt{cos}(5\pi)\:- \texttt{cos}(\pi) \right]\:- \left[ -\frac{1}{5}\texttt{cos}(-5\pi)\:- \texttt{cos}(-\pi)\right] \right)
& \color{gray}{\text{[Evaluate the antiderivative]}} \\[5pt]
&= \frac{1}{2} \left( \left[ -\frac{1}{5}(-1)\:- (-1) \right]\:- \left[ -\frac{1}{5}(-1)\:- (-1)\right] \right) \\[5pt]
&= \frac{1}{2} \left( \left[ \frac{1}{5}\:+ 1 \right]\:- \left[ \frac{1}{5}\:+ 1\right] \right) \\[5pt]
&= \frac{1}{2} \left( \frac{6}{5}\:- \frac{6}{5} \right) \\[5pt]
&= \frac{1}{2} (0) \\[5pt]
&= \color{green}{0}
\end{align*}
\)

Apply the pythagoras theorem: \(\texttt{sin}^2(x)+\texttt{cos}^2(x) = 1\), so \(\texttt{sin}^2(x) = 1\:- \texttt{cos}^2(x)\)

 

\(\normalsize
\begin{align*}\displaystyle
\int^{\frac{\pi}{4}}_0 \texttt{sin}^3(x)dx = \int^{\frac{\pi}{4}}_0 \texttt{sin}(x) \cdot \texttt{sin}^2(x) = \int^{\frac{\pi}{4}}_0 \texttt{sin}(x)[1\:- \texttt{cos}^2(x)]dx
\end{align*}
\)

 

 

Apply a U-Sub. Let \(u=\texttt{cos}(x)\), so \(du=-\texttt{sin}(x)dx\). Substituting these values leads to:

\(\normalsize
\begin{align*}\displaystyle
\int^{\frac{\pi}{4}}_0 [1-\texttt{cos}^2(x)]\texttt{sin}(x)dx
&= \int^{\frac{\pi}{4}}_0 [1-\color{red}{\underbrace{\color{black}{\texttt{cos}^2(x)}}_{\large u^2}}] \color{red}{\underbrace{\color{black}{\texttt{sin}(x)}}_{\large -du}}
& \color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= -\int^{\frac{\pi}{4}}_0 (1-u^2)du
& \color{gray}{\text{[Rearrange the expression]}} \\[5pt]
&= -\int^{\frac{\pi}{4}}_0 du\:+ \int^{\frac{\pi}{4}}_0 u^2du
& \color{gray}{\text{[Separate the integrals]}} \\[5pt]
&= [-u\:+ \frac{1}{3}u^3]\Big|^{\frac{\pi}{4}}_0
& \color{gray}{\text{[Integrate]}} \\[5pt]
&= \left[ \frac{1}{3}\texttt{cos}^3(x)\:- \texttt{cos}(x) \right] \Big|^{\frac{\pi}{4}}_0
& \color{gray}{\text{[Substitute back to x and rearrange]}} \\[5pt]
&= \left[ \frac{1}{3}\texttt{cos}^3(\frac{\pi}{4})\:- \texttt{cos}(\frac{\pi}{4}) \right]\:- \left[ \frac{1}{3}\texttt{cos}^3(0)\:- \texttt{cos}(0) \right]
& \color{gray}{\text{[Evaluate the antiderivative]}} \\[5pt]
&= \left[ \frac{1}{3}\left( \frac{\sqrt{2}}{2} \right)^3\:- \frac{\sqrt{2}}{2} \right]\:- \left[ \frac{1}{3}(1)^3\:- 1 \right] \\[5pt]
&= \left[ \frac{1}{3}\left( \frac{2\sqrt{2}}{8} \right)\:- \frac{\sqrt{2}}{2} \right]\:+\frac{2}{3} \\[5pt]
&= \frac{\sqrt{2}}{12}\:- \frac{\sqrt{2}}{2}\:+\frac{2}{3} \\[5pt]
&= -\frac{5\sqrt{2}}{12}\:+ \frac{2}{3} \\[5pt]
&= \color{green}{\frac{2}{3}\:- \frac{5\sqrt{2}}{12}} \\[5pt]
\end{align*}
\)

Apply a quick U-sub. Let \(u=4x\), so \(du=4dx\). Substituting these values yield:

 

\(\normalsize
\begin{align*}\displaystyle
\int \texttt{sec}(4x)dx
&= \int \texttt{sec}(\color{red}{\underbrace{\color{black}{4x}}_{\large u}}) \color{red}{\underbrace{\color{black}{dx}}_{\large \frac{du}{4}}}
& \color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= \frac{1}{4}\int \texttt{sec}(u)du
& \color{gray}{\text{[Rearrange the expression]}} \\[5pt]
&= \frac{1}{4}\texttt{ln} |\texttt{sec}(u)\:+ \texttt{tan}(u)| \:+ C
& \color{gray}{\text{[Integrate]}} \\[5pt]
&= \color{green}{\frac{1}{4}\texttt{ln}|\texttt{sec}(4x)\:+ \texttt{tan}(4x)|\:+ C}
& \color{gray}{\text{[Substitute back to x]}} \\[5pt]
\end{align*}
\)

Apply a quick U-sub. Let \(u=3x\), so \(du=3dx\). Substituting these values yield:

\(\normalsize
\begin{align*}\displaystyle
\int \texttt{tan}^3(3x) \texttt{sec}^2(3x) dx
&= \int \texttt{tan}^3 \color{red}{\underbrace{\color{black}{(3x)}}_{\large u}} \texttt{sec}^2 \color{red}{\underbrace{\color{black}{(3x)}}_{\large u}} \color{red}{\underbrace{\color{black}{dx}}_{\large \frac{du}{3}}}
& \color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= \frac{1}{3}\int \texttt{tan}^3(u) \texttt{sec}^2(u)du
& \color{gray}{\text{[Simplify]}} \\[5pt]
&= \frac{1}{3}\int \texttt{tan}(u) \color{red}{\underbrace{\color{black}{[\texttt{sec}^2(u)\:- 1]}}_{\texttt{tan}^2(u)}} \texttt{sec}^2(u)du
& \color{gray}{\text{[Apply Pythagoras Theorem]}}
\end{align*}
\)

 

Apply a quick U-sub. Let \(v=\texttt{sec}(u)\), so \(dv=\texttt{sec}(u)\texttt{tan}(u)du\). Substituting these values yield:

\(\normalsize
\begin{align*}\displaystyle
\frac{1}{3} \int [\texttt{sec}^2(u)\:- 1]\texttt{sec}^2(u)\texttt{tan}(u)du
&= \frac{1}{3} \int [\color{red}{\underbrace{\color{black}{\texttt{sec}^2(u)}}_{\large v^2}}\:- 1] [\color{red}{\underbrace{\color{black}{\texttt{sec}(u)}}_{\large v}}] [\color{red}{\underbrace{\color{black}{\texttt{sec}(u)\texttt{tan}(u)du}}_{\large dv}}]
& \color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= \frac{1}{3}\int (v^2\:- 1)vdv
& \color{gray}{\text{[Simplify]}} \\[5pt]
&= \frac{1}{3}\int (v^3\:- v)dv \\[5pt]
&= \frac{1}{3}\left( \frac{1}{4}v^4\:- \frac{1}{2}v^2 \right)\:+ C
& \color{gray}{\text{[Integrate]}} \\[5pt]
&= \frac{1}{12}v^4\:- \frac{1}{6}v^2\:+ C
& \color{gray}{\text{[Simplify]}} \\[5pt]
&= \frac{1}{12}\texttt{sec}^4(u)\:- \frac{1}{6}\texttt{sec}^2(u)\:+ C
& \color{gray}{\text{[Substitute back to u]}} \\[5pt]
&= \color{green}{\frac{1}{12}\texttt{sec}^4(3x)\:- \frac{1}{6}\texttt{sec}^2(3x)\:+ C}
& \color{gray}{\text{[Substitute back to x]}} \\[5pt]
\end{align*}
\)

Apply a quick U-sub. Let \(u=\texttt{tan}(x)\), so \(du=\texttt{sec}^2(x)dx\). Substituting these values yield:

\(\normalsize
\begin{align*}\displaystyle
\int \texttt{tan}^4(x) \texttt{sec}^2(x) dx
&= \int \color{red}{\underbrace{\color{black}{\texttt{tan}^4(x)}}_{\large u^4}} \color{red}{\underbrace{\color{black}{\texttt{sec}^2(x)dx}}_{\large du}}
& \color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= \int u^4du
& \color{gray}{\text{[Simplify]}} \\[5pt]
&= \frac{1}{5} u^5\:+ C
& \color{gray}{\text{[Integrate]}} \\[5pt]
&= \color{green}{\frac{1}{5} \texttt{tan}^5(x)\:+ C}
& \color{gray}{\text{[Substitute back to x]}}
\end{align*}
\)

Use the power reduction identity in order to transform the integrand by parts, first the numerator and then the denominator. Let \(\texttt{sin}^2(x)=\frac{1}{2}[1-\texttt{cos}(2x)]\), so \(2\texttt{sin}^2(x)=1-\texttt{cos}(2x)\). Likewise, let \(\texttt{cos}^2(x)=\frac{1}{2}[1+\texttt{cos}(2x)]\), so \(2\texttt{cos}^2(x)=1+\texttt{cos}(2x)\). Substituting these new values in the integrand yields:

\(\normalsize
\begin{align*}\displaystyle
\int \frac{1+\texttt{cos}(2x)}{1-\texttt{cos}(2x)}dx
&= \int \frac{\color{red}{2\texttt{cos}^2(x)}}{\color{red}{2\texttt{sin}^2(x)}} dx
& \color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= \int \texttt{cot}^2(x)dx
& \color{gray}{\text{[Simplify and use trigonometric identitites]}} \\[5pt]
&= \int [\texttt{cot}^2(x)+1-1]dx
& \color{gray}{\text{[Add zero to the integrand]}} \\[5pt]
&= \int [\texttt{cot}^2(x)+1]dx\:- \int dx
& \color{gray}{\text{[Separate the integrals]}} \\[5pt]
&= \int \texttt{csc}^2(x)dx\:- \int dx
& \color{gray}{\text{[Simplify using trigonometric identitites]}} \\[5pt]
&= \color{green}{\texttt{cot}(x)\:-x\:+C}
& \color{gray}{\text{[Integrate]}} \\[5pt]
\end{align*}
\)

Multiply the integrand by its reciprocal in a fraction:

\(\normalsize
\begin{align*}\displaystyle
\int \sqrt{1-\texttt{cos}(\theta)} d\theta
&= \int \sqrt{1-\texttt{cos}(\theta)} \left( \frac{\sqrt{1+\texttt{cos}(\theta)}}{\sqrt{1+\texttt{cos}(\theta)}} \right) d\theta \\[5pt]
&= \int \frac{\sqrt{1-\texttt{cos}^2(\theta)}}{\sqrt{1+\texttt{cos}(\theta)}} d\theta
& \color{gray}{\text{[Simplify]}} \\[5pt]
&= \int \frac{\sqrt{\texttt{sin}^2(\theta)}}{\sqrt{1+\texttt{cos}(\theta)}} d\theta \\[5pt]
&= \int \frac{\texttt{sin}(\theta)}{\sqrt{1+\texttt{cos}(\theta)}} d\theta \\[5pt]
\end{align*}
\)


Perform a quick U-Sub. Let \(u=1\:+ \texttt{cos}(\theta)\), so \(du=-\texttt{sin}(\theta)d\theta\):

\(\normalsize
\begin{align*}\displaystyle
\int \frac{\texttt{sin}(\theta)}{\sqrt{1+\texttt{cos}(\theta)}} d\theta
&= \int \frac{\color{red}{-du}}{\color{red}{\sqrt{u}}}
& \color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= -2\sqrt{u}\:+ C
& \color{gray}{\text{[Integrate]}} \\[5pt]
&= \color{green}{-2\sqrt{1\:+ \texttt{cos}(\theta)}\:+ C}
& \color{gray}{\text{[Substitute back for \(\theta\)]}} \\[5pt]
\end{align*}
\)

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