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\leftline{MATH 136, section 1 -- Calculus 2}
\leftline{Integration By Substitution Practice -- Solutions}
\leftline{February 7, 2014}
\vskip 20pt
\noindent
Each of the following integrals can be done by the method of $u$-substitution:
\begin{enumerate}
\item Determine the appropriate function $u$, 
\item Compute $du$, 
\item Change the integral to an equivalent form in the new variable $u$.  If
it is a definite integral, you can convert the limits of integration as well.
\item Integrate, then 
\item Resubstitute $u$ to express the answer in terms of the original variable (indefinite
integral cases), or evaluate (definite integral cases).
\end{enumerate}
\bigskip
\begin{itemize}
\item $\displaystyle \int x\sqrt{x^2 + 16}\ dx$.  Let $u = x^2 + 16$.  Then 
$du = 2x\ dx$ and $x\ dx = \frac{1}{2} du$.  So the integral is
$$\frac{1}{2} \int \sqrt{u}\ dx = \frac{1}{2}\cdot \frac{2}{3} u^{3/2} + C = 
\frac{1}{3}(x^2 + 16)^{3/2} + C.$$ 
\item $\displaystyle \int \cos\theta e^{1 + \sin\theta}\ d\theta$.  Let $u = 1 + \sin\theta$.
Then $du = \cos\theta$ and the integral becomes:
$$\int e^u\ du = e^u + C = e^{1 + \sin\theta} + C.$$
\item $\displaystyle \int_{\pi/4}^{\pi/2} \cos^3(4\theta) \sin(4\theta)\ d\theta$.
Let $u = \cos(4\theta)$.  Then $du = -4\sin(4\theta)\ d\theta$ (chain rule) and 
$\sin(4\theta)\ d\theta = \frac{-1}{4}du$.  Also, 
the limits of integration become $x = \pi/4 \Rightarrow u = \cos(\pi) = -1$
and $x = \pi/2 \Rightarrow u = \cos(2\pi) = 1$.  So the integral becomes
$$\frac{-1}{4} \int_{-1}^{1} u^3\ du = \left. \frac{-u^4}{16} \right|_{-1}^1 = 0.$$
\item $\displaystyle \int \frac{1}{\sqrt{1 - 4x^2}}\ dx$.  For this one, note
the similarity with the formula for the derivative of the inverse sine function
if we think of $\sqrt{1 - 4x^2}$ as $\sqrt{1 - (2x)^2}$.
If $u = 2x$, then $du = 2\ dx$ so the integral becomes:
$$\frac{1}{2}\int \frac{du}{\sqrt{1 - u^2}} = \frac{1}{2}\sin^{-1}(u) + C 
= \frac{1}{2}\sin^{-1}(2x) + C.$$
\item $\displaystyle \int_0^{1/4} \frac{x}{\sqrt{1 - 4x^2}}\ dx$.  This one is 
different from the previous one because of the extra $x$ in the numerator.
That $x$ is part of $du$ for $u = 1 - 4x^2$.  With that substitution, the 
integral becomes 
$$\frac{-1}{8} \int_{u=1}^{u=3/4} u^{-1/2}\ du = \left. \frac{-1}{4} u^{1/2}\right|_{1}^{3/4} = \frac{1}{4}\left(1 - \sqrt{\frac{3}{4}}\right).$$
\item $\displaystyle \int \frac{1}{x\ln(x)}\ dx$.  Let $u = \ln(x)$, then 
$du = \frac{1}{x}\ dx$ and the integral becomes
$$\int \frac{1}{u}\ du = \ln|u| + C = \ln|\ln(x)| + C.$$
\item $\displaystyle \int \frac{\cos(\sqrt{y})}{\sqrt{y}}\ dy$.  Let $u = \sqrt{y}$.
Then $du = \frac{1}{2\sqrt{y}}\ dy$, so the integral is
$$2\int \cos(u)\ du = 2\sin(u) + C = 2\sin(\sqrt{y}) + C.$$
\end{itemize}
\end{document}
