Practice Problems: Partial Fractions

Apply Weierstrass substitution. Let \(\texttt{sin}(\theta)=\frac{2x}{1+x^2}\), \(\texttt{cos}(\theta)=\frac{1-x^2}{1+x^2}\), \(x=\texttt{tan}(\frac{\theta}{2})\), and \(d\theta = \frac{2dx}{1+x^2}\). Applying this substitution in the integrand leads to:

\(
\begin{align*}
\int \frac{\color{red}{\texttt{sin}(\theta)}}{\color{red}{\texttt{sin}(\theta)} + \color{red}{\texttt{cos}(\theta)}} \color{red}{d\theta}
&= \int \frac{\color{red}{\frac{2x}{1+x^2}}}{\color{red}{\frac{2x}{1+x^2}} + \color{red}{\frac{1-x^2}{1+x^2}}} \color{red}{\left( \frac{2}{1+x^2}dx \right)}
& \color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= 4 \int \frac{\frac{x}{(1+x^2)^2}}{\frac{-x^2+2x+1}{1+x^2}}dx
& \color{gray}{\text{[Manipulate algebraically]}} \\[5pt]
&= 4 \int \left( \frac{x}{(1+x^2)^2} \right) \left( \frac{1+x^2}{-x^2+2x+1} \right) dx
& \color{gray}{\text{[Reverse the denominator]}} \\[5pt]
&= -4 \int \frac{x}{(1+x^2)(x^2-2x-1)}dx
& \color{gray}{\text{[Simplify]}} \\[5pt]
\end{align*}
\)

In order to apply partial fractions decomposition use the quadratic formula for the second factor in the denominator:

\(
\begin{align*}
& x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} = \frac{-(-2) \pm \sqrt{(-2)^2-4(1)(-1)}}{2(1)} = \frac{2 \pm \sqrt{4+4}}{2} = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2} \\[5pt]
& x_1 = 1 + \sqrt{2} \\[5pt]
& x_2 = 1 – \sqrt{2} \\[5pt]
& x^2-2x-1 = (x – 1 + \sqrt{2})(x – 1 – \sqrt{2})
\end{align*}
\)

Update the factor in the denominator with the new factorization result to apply partial fractions decomposition:

\(
\begin{align*}
&-4\int \frac{x}{(1+x^2)\color{red}{(x^2-2x-1)}}dx = -4\int \frac{x}{(1+x^2)\color{red}{(x-1+\sqrt{2})(x-1-\sqrt{2})}}dx \\[5pt]
&= -4\left( \int \frac{Ax+B}{1+x^2}dx + \int \frac{C}{x-1+\sqrt{2}}dx + \int \frac{D}{x-1-\sqrt{2}}dx \right) \\\\
& \color{gray}{\text{[Use the equating method]}} \\[5pt]
&x = (Ax+B)(x-1+\sqrt{2})(x-1-\sqrt{2}) + \color{blue}{C(x^2+1)(x-1-\sqrt{2})} + \color{red}{D(x^2+1)(x-1+\sqrt{2})} \\[5pt]
& x = Ax^3-2Ax^2-Ax+Bx^2-2Bx-B + \color{blue}{Cx^3-Cx^2-\sqrt{2}Cx^2+Cx-C-\sqrt{2}C} + \color{red}{Dx^3-Dx^2+\sqrt{2}Dx^2+Dx-D+\sqrt{2}D} \\[5pt]
& x = \color{red}{[A+C+D]}x^3 + \color{blue}{[-2A+B-C(1+\sqrt{2})-D(1-\sqrt{2})]}x^2 + \color{green}{[-A-2B+C + D]}x + \color{orange}{[-B-C(1+\sqrt{2})-D(1-\sqrt{2})]} \\\\
& \color{gray}{\text{[Solve the resulting equation system using any method]}} \\[5pt]
&A+C+D=0 \\[5pt]
&-2A+B-C-\sqrt{2}C-D+\sqrt{2}D=0 \\[5pt]
&-A-2B+C+D=1 \\[5pt]
&-B-C-\sqrt{2}C-D+\sqrt{2}D=0 \\[5pt]
& A = -\frac{1}{4}\\[5pt]
& B = -\frac{1}{4} \\[5pt]
& C = \frac{1}{8} \\[5pt]
& D = \frac{1}{8} \\[5pt]
\end{align*}
\)

Update the newfound values into the partial fraction integrals:

\(
\begin{align*}
-4&\int \frac{x}{(1+x^2)(x-1+\sqrt{2})(x-1-\sqrt{2})}dx\\
&= -4\left( \int \frac{-\frac{1}{4}x-\frac{1}{4}}{1+x^2}dx + \int \frac{\frac{1}{8}}{x-1+\sqrt{2}}dx + \int \frac{\frac{1}{8}}{x-1-\sqrt{2}}\right)
& \color{gray}{\text{[Update]}} \\[5pt]
&= \int \frac{x+1}{1+x^2}dx\:- \frac{1}{2}\int \frac{1}{x-1+\sqrt{2}}dx\:- \frac{1}{2}\int \frac{1}{x-1-\sqrt{2}}dx \\[5pt]
&= \int \frac{x}{1+x^2}dx + \int \frac{1}{1+x^2}dx\: – \frac{1}{2} \left( \texttt{ln}|x-1+\sqrt{2}|\:+ \texttt{Ln}|x-1-\sqrt{2}| \right) + C
& \color{gray}{\text{[Integrate]}} \\[5pt]
& = \frac{1}{2}\int \frac{1}{u}du + \int \frac{1}{1+x^2}dx\: – \frac{1}{2} \texttt{ln}|(x-1+\sqrt{2})(x-1-\sqrt{2})| + C \\[5pt]
& = \frac{1}{2} \texttt{ln}|u| + \texttt{arctan}(x)\:- \frac{1}{2} \texttt{ln}|x^2-2x-1| + C \\[5pt]
& = \frac{1}{2} \texttt{ln}(x^2+1)\:- \frac{1}{2} \texttt{ln}|x^2-2x-1| + \texttt{arctan}(x) + C \\[5pt]
&= \frac{1}{2} \texttt{ln} \left| \frac{x^2+1}{x^2-2x-1} \right| +\texttt{arctan}(x) + C
& \color{gray}{\text{[Substitute back]}} \\[5pt]
&= \color{green}{\frac{1}{2} \texttt{ln} \left| \frac{\texttt{tan}^2(\frac{\theta}{2})+1}{\texttt{tan}^2(\theta)-2\texttt{tan}(\frac{\theta}{2})-1} \right| + \texttt{arctan} \left( \texttt{tan} \left( \frac{\theta}{2} \right) \right) + C}
\end{align*}
\)