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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 method of $u$-substitution is a \emph{change of variables}
method, where we essentially rewrite the integral above as 
$$\int f(u)\ du = F(u) + C$$
involving functions of only the new variable $u$.  The goal is to use
the known antiderivative $F(u)$ for the function $f(u)$ as a function of $u$.

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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[] {\it Solution:}  Let $u = x^3$, then $du = 3x^2\ dx$, so the $x^2 = \frac{1}{3}du$
and the integral becomes 
$$\int \frac{1}{3}\cos(u)\ du = \frac{1}{3} \sin(u) + C = \frac{1}{3} \sin(x^3) + C$$
\item $\displaystyle \int (\sqrt{x} + 1)^4\cdot \frac{1}{\sqrt{x}}\ dx$
\item[] {\it Solution:}  Let $u = \sqrt{x} + 1 = x^{1/2} + 1$. Then $du = \frac{1}{2} x^{-1/2}\ dx = \frac{dx}{2\sqrt{x}}$.
In terms of $u$, the integrand becomes:
$$\int u^4 \cdot 2\ du = \frac{2}{5} u^5 + C = \frac{2}{5}(\sqrt{x} + 1)^5 + C$$ 
\item[] $\displaystyle \int \frac{x + 4}{x^2 + 8x + 9}\ dx$ 
\item {\it Solution:}  Here we have to recognize that if $u = x^2 + 8x + 9$, then $du = (2x + 8)\ dx = 2(x + 4)\ dx$.
This is twice the numerator of the fraction so the form is 
$$\frac{1}{2} \int \frac{1}{u}\ du = \frac{1}{2} \ln|u| + C =  \frac{1}{2} \ln|x^2 + 8x + 9| + C$$ 
\item $\displaystyle \int e^{\tan(x)} \sec^2(x)\ dx$
\item[] {\it Solution:}  Let $u = \tan(x)$.  Then $du = \sec^2(x)\ dx$ on the nose.  So we
have $\int e^u\ du = e^u + C = e^{\tan(x)} + C$.
\item $\displaystyle \int \frac{x}{\sqrt{1 - x^4}}\ dx$
\item[] {\it Solution:}  This looks like an inverse sine integral.  To verify that, 
notice that $x^4 = (x^2)^2$.  So with $u = x^2$, we have $du = 2x\ dx$, and the integrand is
$$\frac{1}{2} \int \frac{du}{\sqrt{1 - u^2}} = \frac{1}{2}\sin^{-1}(u) + C = \frac{1}{2} \sin^{-1}(x^2) + C$$
\end{enumerate}

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