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\title{Math 1a.  Lecture 3 \\
Integration by Parts}

\author{T. Judson}

\date{Spring 2006}


\begin{document}

\maketitle


\section{Goals}

\begin{itemize}

\item
To be able to evaluate definite and indefinite integrals using integration by parts.


\end{itemize} 



\section{Integration by Parts}

\begin{itemize}

\item
$\displaystyle \int u \, dv = uv - \int v \, du$ 

\item
$\displaystyle \int_a^b u \, dv =  uv \bigg]_a^b - \int_a^b v \, du$ 


\end{itemize}
Since
$$
(u \cdot v)' = u \cdot v' + u' \cdot v.
$$
we can integrate both sides to obtain
$$
\int u \, dv = uv - \int v \, du.
$$




\section{Key Examples}

\begin{multicols}{2}
\begin{itemize}


\item 
$\displaystyle \int x e^x \, dx$ and $\displaystyle \int_0^1 x e^x \, dx$

\item 
$\displaystyle \int x^2 e^x \, dx$


\item 
$\displaystyle \int \ln x  \, dx$

\item 
$\displaystyle \int e^x \cos x \, dx$




\end{itemize}
\end{multicols}

\section{LIATE}

Students often ask how should I choose $u$ and $dv$.  The answer is {\em LIATE}.  That is, choose $u$ in the following order:
\begin{itemize}

\item
{\bf L}ogarithmic

\item
{\bf I}nverse

\item
{\bf A}lgebraic

\item
{\bf T}rigonometric

\item
{\bf E}xponential

\end{itemize}
For example, let $u = \arctan x$ and $dv = dx$ in $\int \arctan x \, dx$, since $\arctan x$ is an {\bf I}nverse function.  Then
\begin{eqnarray*}
\int \arctan x \, dx & = & uv - \int v \, du \\
& = & 
x \arctan x - \int \frac{x}{1 + x^2} \, dx \\
& = &  
x \arctan x - \frac{1}{2} \ln(1 + x^2).
\end{eqnarray*}


\section*{References}

\begin{itemize}

\item
\S 5.6 in 
James Stewart.
{\it Single Variable Calculus: Concepts \& Context}, third edition.
Brooks/Cole, Belmont CA, 2005.
ISBN 0--534--41022--7.


\end{itemize}



\section*{Notes}

\subsection*{February 8, 2006}



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