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\leftline{MATH 132, section 1 -- Calculus for Physical and Life Sciences}
\leftline{Discussion 2 -- Integration By Parts Practice}
\leftline{February 4, 2008}
\vskip 20pt
\noindent
The method of \emph{integration by parts}, applies the 
``reverse product rule''
\begin{equation}\label{rpr}
\int u\ dv = u v - \int v\ du
\end{equation}
to ``trade in'' a harder integral $\int u\ dv$ for a 
hopefully simpler one $\int v\ du$.  The steps for this method are 
\begin{enumerate}
\item Split up the original integrand into the two parts $u$ and $dv$. (Note:
everything has to go into one part or the other -- no ``left-overs'')
\item From $u$ compute $du = u'(x) dx$.  From $dv$, compute $v = \int dv$.  
\item Apply the reverse product rule (\ref{rpr})
\item Finish by computing $\int v\ du$.  (This could require a substitution, 
or some other method.)
\item When there is a choice for the $u$ and $dv$, the good choice
will give $\int v\ du$ that is simpler (or at least no more complicated
than) $\int u\ dv$.  If your $\int v\ du$ is more complicated (cannot 
be simplified), then you might need to reconsider the choice of $u$, $dv$.
\end{enumerate}
Do all the integrals below, and put the work for your group's integral up 
on the board.
\bigskip
\begin{enumerate}
\item (I) and (V) \[ \displaystyle \int x^2 \cos(3x)\ dx\]
\vskip 15pt
\item (II) and (VI) \[\displaystyle \int x^3\sqrt{9 + x^2}\ dx\]
(Hint: The $dv$ for this one should contain the $\sqrt{9 + x^2}$.  
\emph{What else} do you need to be able to integrate $dv$
by a method we have discussed?)
\vskip 15pt
\item (III) and (VII) \[\displaystyle \int \tan^{-1}{3x} \ dx\]
\vskip 15pt
\item (IV) and (VIII)  \[\displaystyle \int x\ln(x + 2)\ dx\] 
(Hint: Substitute $w = x + 2$, then use parts.)
\end{enumerate}
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