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\centerline{MATH 134 -- Calculus with Fundamentals 2}
\centerline{Practice Day on integration ''by parts''}
\centerline{February 28, 2018}
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\noindent
{\it Background}
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The integration by parts method is based on the product rule for derivatives, 
but we usually apply it by remembering the ``parts formula:"
$$\int u\ dv = uv - \int v\ du.$$
The idea is that we break up an integral we want to compute into a $u$ and a $dv$.
We compute $du = \frac{du}{dx}\ dx$ and $v = \int dv$ and apply the parts formula.
\emph{We then have to finish by integrating} $\int v\ du$.   
If we have made a good choice, then the integral $\int v\ du$ is simpler, or at least
no harder than the integral we started from.  
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{\it Questions}
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\noindent
Integrate by parts
\begin{enumerate}
\item $\displaystyle \int x e^{2x}\ dx$
\item $\displaystyle \int x^2 \cos(\pi x)\ dx$
\item $\displaystyle \int x^7 \ln(3x)\ dx$
\item $\displaystyle \int \sin^{-1}(x)\ dx$
\item A more challenging one:  $\displaystyle \int e^{3x} \sin(5x)\ dx$.  Hint: you'll
need to integrate by parts twice, and it will seem as though you are going in circles.
But be of good cheer and persist -- things will work out!
\item Check by differentiating (and simplifying!):
$$\int \sec(x)\ dx = \ln|\sec(x) + \tan(x)| + C$$
\item Another challenging one:  $\displaystyle \int \sec^3(x)\ dx$.  Hint:
Try making $u = \sec(x)$ and $dv = \sec^2(x)\ dx$.  You'll also need the trig identity 
$1 + \tan^2(x) = \sec^2(x)$ and the previous formula for $\int \sec(x)\ dx$.  
\end{enumerate}
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\caption{The region for problem 4.}
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