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\begin{document}


\title{On-line Math 21}


\email{david.johnson@lehigh.edu}


\urladdr{http://www.lehigh.edu/\textasciitilde{}dlj0}

\maketitle

\subsection{Techniques of integration}

\begin{example}
\[
\int e^{x}\sin (x)dx\]

\end{example}

\subsubsection*{Solution}

You apply parts twice to this one, too, but in a different way. First, set \( u=e^{x} \),
and \( dv=\sin (x)dx \), giving the following table:

\vspace{0.3cm}
{\centering \begin{tabular}{|c|c|}
\hline 
\( u=e^{x} \)&
\( du=e^{x}dx \)\\
\hline 
\( v=-\cos (x) \)&
\( dv=\sin (x)dx \)\\
\hline 
\end{tabular}\par}
\vspace{0.3cm}

Then,\begin{eqnarray*}
\int e^{x}\sin (x)dx & = & \int u\, dv\\
 & = & u\, v-\int v\, du\\
 & = & e^{x}(-\cos (x))-\int (-\cos (x))e^{x}dx\\
 & = & -e^{x}\cos (x)+\int e^{x}\cos (x)dx,
\end{eqnarray*}
which seems to be spinning our wheels, leading us back essentially where we
started. Now we do it again, keeping the choice of \( u \) to be the exponential
term, and \( dv \) the trigonometric:

\vspace{0.3cm}
{\centering \begin{tabular}{|c|c|}
\hline 
\( u=e^{x} \)&
\( du=e^{x}dx \)\\
\hline 
\( v=\sin (x) \)&
\( dv=\cos (x)dx \)\\
\hline 
\end{tabular}\par}
\vspace{0.3cm}

Then, applying parts to the integral term left over from the first application,
\begin{eqnarray*}
\int e^{x}\sin (x)dx & = & -e^{x}\cos (x)+\int e^{x}\cos (x)dx\\
 & = & -e^{x}\cos (x)+\left( e^{x}\sin (x)-\int \sin (x)e^{x}dx\right) \\
 & = & -e^{x}\cos (x)+e^{x}\sin (x)-\int e^{x}\sin (x)dx.
\end{eqnarray*}
You'll notice that we have now come full circle. But there is a twist. We did
not come back to exactly the same term; there is a (vital) negative sign there,
and also there are those other terms. Look from the first line in that long
equation, to the last:\[
\int e^{x}\sin (x)dx=-e^{x}\cos (x)+e^{x}\sin (x)-\int e^{x}\sin (x)dx.\]
You can solve that equation for \( \int e^{x}\sin (x)dx \), which is, after
all, what we are trying to do. Mysteriously, when you get all integrals on one
side, there should be an undetermined constant on the other, reflecting the
fact that the integral involves an unknown constant term: \[
2\int e^{x}\sin (x)dx=-e^{x}\cos (x)+e^{x}\sin (x)+C,\]
or\[
\int e^{x}\sin (x)dx=\frac{1}{2}e^{x}\left( \sin (x)-\cos (x)\right) +C.\]
There is no need to divide the \( +C \) by 2, since after all it is an undetermined
constant. It is just as undetermined after you divide it by two.
\end{document}
