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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}

In general, finding integrals is a much more complicated process than finding
derivatives. The straightforward algorithms of differentiation may take some
time to be natural for you, but you never have a situtation where there is no
algorithm. With integration, though, it is not uncommon to have to guess which
method to use, to have to try several methods before you find the right one,
and even to find that no method will work.

We deal here with the two most important methods of integration, substitution
and integration by parts. These techniques, even in this day of computers that
can integrate most functions you might need, are still useful, since they do
more than provide algorithms to find antiderivatives. They are general techniques
to decompose problems into rational components. 

There are other techniques of integration that you will learn about in subsequent
courses: partial fractions (linked here to a brief, optional introduction),
trigonometric substitutions, and various tricks. These techniques combine the
two methods we discuss here with algebra and trigonomety to produce refined
techniques that apply in specific situations. These two techniques, however,
are the basis for all the others.


\subsubsection{Substitution}

The point of substitution is to run the chain rule backwards. The chain rule
says that, if \( F'(x)=f(x) \), then \[
(F(u(x))'=F'(u(x))\, u'(x)=f(u(x))\, u'(x),\]


or, in Leibniz notation, \[
\frac{d}{dx}(F(u(x))=f(u(x))\frac{du}{dx}.\]


That last version is the key to substitution. Then, \textit{if} we want to integrate
\emph{exactly} the expression: \[
\int f(u(x))\frac{du}{dx}dx,\]
 the integral would be \( F(u(x))+C \).

Now, if we think about \( u \) as a \textit{variable}, \( u(x)=u \), and we
blithely use algebra on differentials, so that \[
\frac{du}{dx}dx=du,\]
 we can write the line above by: \[
\int f(u)du,\]
 which is \( F(u) \). But, for this last case, you forgot that \( x \) was
the variable at all. \( u \) became the variable, and you just were asked for
the antiderivative of \( f \), which is just \( F \). The trick is in the
fine print: the switch of the differentials. This is the whole reason for the
\( dx \) at the end. That way, all you have to do to \textit{correctly} substitute
is to, if you can, turn everything, \textit{including} the \( dx \), into \( u \)'s,
using the formula for \( u(x) \) and the equation \[
\frac{du}{dx}dx=du.\]
 Then integrate the expression involving \( u \), and substitute back. 

\begin{example}
\[
\int x\sqrt{x^{2}+4}dx.\]
 
\end{example}

\subsubsection*{Solution}

\begin{example}
\[
\int \frac{x^{2}}{x^{3}+7}dx\]
 
\end{example}

\subsubsection*{Solution}

\begin{xca}
\[
\int \sin ^{2}(x)\cos (x)dx=\]
 
\end{xca}
%

\begin{xca}
\[
\int e^{3x+1}dx=\]
 
\end{xca}
%

\begin{xca}
\[
\int \frac{x}{x^{4}+1}dx=\]
 
\end{xca}
\emph{Hint:} For this one, the standard substitution of the whole denominator
will not work. Look back over the standard list of integrals to see which one
this can be made to resemble. That will give you the hint you need to find what
substitution to make.

For definite integrals, keeping in the same spirit, you have to \textit{change}
the limits of integration into the \( u \) equivalents, too. Alternately, you
could do the indefinite integration, substitute back in terms of \( x \), and
evaluate the definite integral using that function \textit{of} \( x \) and
the \( x \)-limits of integration.  

\begin{example}
\[
\int _{0}^{4}\sqrt{3x+4}\, dx.\]

\end{example}

\subsubsection*{Solution}

\begin{xca}
\[
\int _{0}^{2}\frac{x^{2}}{x^{3}+7}dx=\]
 
\end{xca}
%

\begin{xca}
\[
\int _{-1}^{1}x^{2}\sin (x^{3})dx=\]
 
\end{xca}
%

\begin{xca}
\[
\int _{-2}^{2}x\sqrt{4-x^{2}}dx=\]

\end{xca}
\emph{Hint:} Something odd happens with this one to the limits of integration
when you do the substitution. 


\subsubsection{Integration by parts}

This technique is the most powerful in terms of applications beyond mere computing
of integrals. It's absolutely vital to understanding much of differential equations. 

The idea is sort of ``robbing Peter to pay Paul''. We are going to try to run
the product rule backwards. However, the right-hand side has two pieces, \[
(u\, v)'=u'\, v+u\, v',\]


so it's a little hard to do that. We can, however, integrate both sides, which
gives:\[
\int (u\, v)'dx=\int u'\, vdx+\int u\, v'dx,\]
which doesn't look like much improvement. However, you can do the left-hand
side, it's just \( u\, v \). Also, you can solve for \textit{one} of the terms
on the right-hand side. We're going to pick the second one. Solving for that
term,\[
\int u\, v'dx=u\, v-\int u'\, vdx.\]
Now, that's it, really. One integral, \( \int u\, v'dx \), is expressed in
terms of the other, \( \int u'\, vdx \). If you have an integral you can think
of as the integral of a product, and you can pick which piece you want to be
\( u \) and which piece \( v' \), then you can trade that integral in for
the integral of \( v\, u' \). 

We usually write this, like we did for substitution, in differential notation,
where \( du=u'dx \) and \( dv=v'dx \). This then is integration by parts as
you should remember it:\[
\int u\, dv=u\, v-\int v\, du.\]
Let's try a simple example: 

\begin{example}
\[
\int xe^{x}dx\]

\end{example}

\subsubsection*{Solution}

\begin{example}
\[
\int x^{2}\sin (x)dx.\]
 
\end{example}

\subsubsection*{Solution}

\begin{xca}
\[
\int x\ln (x)dx=\]

\end{xca}
%

\begin{xca}
\[
\int \ln (x)dx=\]

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

\end{example}

\subsubsection*{Solution}

If you have an integration by parts problem with limits of integration, be sure
to evaluate the \( u\, v \) part: \[
\int _{a}^{b}u\, dv=\left. u\, v\right| _{a}^{b}-\int _{a}^{b}v\, du.\]



\subsubsection{Partial fractions}
\end{document}
