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\leftline{MATH 136 -- Calculus 2}
\leftline{Integration By Substitution Practice -- Answers}
\leftline{September 16, 2016}
\vskip 20pt
\begin{itemize}
\item $\displaystyle \int x\sqrt{4 x^2 + 16}\ dx$.  Let $u = 4x^2 + 16$; Answer $= \frac{1}{12}(4x^2+16)^{3/2} + C$.
\vskip 10pt
\item $\displaystyle \int_0^\pi \cos\theta e^{1 + \sin\theta}\ d\theta$. Let $u = 1 + \sin\theta$; Answer $= 0$. 
\vskip 10pt
\item $\displaystyle \int_{\pi/4}^{\pi/2} \cos^3(4\theta) \sin(4\theta)\ d\theta$.
Let $u = \cos(4\theta)$; Answer $= 0$.
\vskip 10pt
\item $\displaystyle \int \frac{1}{\sqrt{1 - 4x^2}}\ dx$.
Let $u = 2x$; Answer $= \frac{1}{2}\sin^{-1}(2x) + C$.
\vskip 10pt
\item $\displaystyle \int \frac{x}{\sqrt{1 - 4x^2}}\ dx$.  Note the difference with the previous one! Let $u = 1 - 4x^2$, Answer $=-\frac{1}{4}(1 - 4 x^2)^{1/2} + C$.
\vskip 10pt
\item $\displaystyle \int \frac{1}{x(\ln(x))^p}\ dx$.  The number $p > 0$ is a constant; your answer should depend on $p$ and there should be two separate cases:  Let $u = \ln(x)$, so the integral is
$$\frac{(\ln(x))^{-p+1}}{-p+1} + C$$
if $p \ne 1$ and 
$$
\ln(\ln(x)) + C
$$ 
if $p = 1$.
\vskip 10pt
\item $\displaystyle \int x^3\sec(x^4)\tan(x^4)\ dx$.
Let $y = x^4$.  The answer is $\frac{1}{4}\sec(x^4) + C$.
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\item $\displaystyle \int \frac{\cos(\sqrt{y})}{\sqrt{y}}\ dy$.
Let $u = \sqrt{y}$.  The answer is $= 2\sin(\sqrt{y}) + C$. 
\vskip 10pt
\item $\displaystyle \int x^3\sqrt{4 x^2 + 16}\ dx$.  More challenging!  Letting $u = 4x^2 + 16$, the
$x^3 = \frac{u - 16}{4} \cdot \frac{1}{8} du$ and the integral is 
$$ = \frac{1}{80}(4x^2 + 16)^{5/2} - \frac{1}{3}(4x^2 + 16)^{3/2} + C.$$
\end{itemize}
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