%134 Worksheet
%trig & trig substitution

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{\bf Integration Techniques Quiz}
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{\large

1.use $u=\sin \pi x \Rightarrow du = \pi \cos \pi x \, dx$
\[
    \int \sin^2 \pi x \cos^5 \pi x \, dx = \int \frac{1}{\pi}u^2(1-u^2)^2 \, du 
	= \frac{1}{\pi} \int u^2 -2u^4+u^6 \, du
\]
\[
	\frac{1}{3\pi}\sin^3 \pi x - \frac{2}{5\pi} \sin^5 \pi x +\frac{1}{7\pi}\sin^7 \pi x + C
\]

\vfill

2.use $w = 2x+1$  (You don't have to, but the problem turns out a lot cleaner this way)
\[
    \int \ln (2x+1) \, dx = \int \frac{1}{2}\ln w \, dw 
\]
\[
\begin{array}{ll}
	u=\ln w & dv=\frac{1}{2}\, dw\\
	du = \frac{1}{w}\, dw & v = \frac{1}{2}w\\
\end{array}
\]
\[
	\int \frac{1}{2}\ln w \, dw 
	= \frac{1}{2}w \ln w - \int \frac{1}{w}\cdot \frac{1}{2}w \, dw 
	= \frac{1}{2}w \ln w - \int \frac{1}{2} \, dw
\]
\[
	=\frac{1}{2}w \ln w - \frac{1}{2}w + C
\]
\[
	=\frac{1}{2}(2x+1)\ln (2x+1) - \frac{1}{2}(2x+1) + C
\]
\\

\vfill

3. use $x=4\tan \theta \rightarrow dx = 4\sec^2 \theta \, d\theta$
\[
    \int \frac{1}{\sqrt{x^2+16}} \, dx 
	= \int \frac{1}{\sqrt{16\sec^2 \theta}} (4\sec^2 \theta)\, d\theta
\]
\[
	\int \sec \theta \, d\theta = \ln | \sec \theta + \tan \theta | + C
\]
\[
	= \ln \Big| \frac{\sqrt{x^2+16}}{4}+\frac{x}{4} \Big| + C
\]
\\

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\[
	\int \frac{1}{(x + 5)^2(x - 1)} \, dx
\]

work to pull rational function apart
\[
	\frac{1}{(x + 5)^2(x - 1)} = \frac{A}{x-1}+\frac{B}{x+5}+\frac{C}{(x+5)^2}
\]
\[
	1=A(x+5)^2+B(x+5)(x-1)+C(x-1) 
\]
\[
	1= (A+B)x^2 +(10A+4B+C)x +25A-5B-C
\]
\[
	\mbox{when } x=1 \Longrightarrow 1 = 36A \Longrightarrow A = \frac{1}{36} \mbox{ and } B=-\frac{1}{36}
\]
\[
	\mbox{when } x=-5 \Longrightarrow 1 = 4C \Longrightarrow \frac{1}{4}
\]
\[
	\int \frac{1}{(x + 5)^2(x - 1)} \, dx 
	= \int \frac{1}{36}\cdot \frac{1}{x-1}-\frac{1}{36}\cdot \frac{1}{x+5}+\frac{1}{4}\cdot\frac{1}{(x+5)^2} \, dx
\]
\[
	= \frac{1}{36}\ln | x-1 | - \frac{1}{36}\ln |x+5| - \frac{1}{4(x+5)} + C
\]




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