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\centerline{MATH 136 -- Calculus 2}
\centerline{Practice Day on $u$-substitution}
\centerline{February 4, 2020}
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\noindent
{\it Background}
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Recall that $u$-substitution is the name given to the indefinite integration
method coming from the Chain Rule for derivatives.  In its most basic form, 
it just says that if $F$ is an antiderivative of $f$, then 
$$\int f(u(x)) \frac{du}{dx}\ dx = F(u(x)) + C.$$  
Note that $\frac{du}{dx} = u'(x)$ appears in the integral on the 
left.  If that term were not there, then this method would not 
``work.''

The most important aspects to master
here are
\begin{itemize}
\item recognizing good candidates for the $u$ to substitute for
\item always remembering to compute $du = \frac{du}{dx}\ dx$ and to
match that with the rest of the integral other than the $f(u)$ part,
\item using the other basic derivative rules to find $F$ from $f$.
\end{itemize}
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\noindent
{\it Questions}

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For each problem, 
\begin{enumerate}
\item[(i)] find a candidate $u$,
\item[(ii)]  compute $du = \frac{du}{dx}\ dx$
\item[(iii)]  see whether the rest of the integrand can be matched with $du$, 
possibly up to a constant multiple (if not, then you might need to try a 
different $u$ or a different method entirely), 
\item[(iv)] finish the integration.
\end{enumerate}
\begin{enumerate}
\item $\displaystyle \int x^2 \cos(x^3)\ dx$
\item $\displaystyle \int (\sqrt{x} + 1)^4\cdot \frac{1}{\sqrt{x}}\ dx$
\item $\displaystyle \int \frac{x + 4}{x^2 + 8x + 9}\ dx$ 
\item $\displaystyle \int e^{\tan(x)} \sec^2(x)\ dx$
\item $\displaystyle \int \frac{x}{\sqrt{1 - x^4}}\ dx$
\end{enumerate}

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