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

\subsubsection*{Solution}

I would take \( u(x) \) to be \( u(x)=x^{2}+4 \). I'm looking for an ``inside
part'' (remember the chain rule), and I \textit{hope} I can find the essential
part of its derivative to use to make \( du \). But, actually, that last part
is fairly automatic. Note that \[
\frac{du}{dx}=2x,\]
 so \( du=2xdx \), or \( du/2x=dx \):\begin{eqnarray*}
\int x\sqrt{x^{2}+4}dx & = & \int x\sqrt{u}du/2x\\
 & = & \frac{1}{2}\int \sqrt{u}du\\
 & = & \frac{1}{2}u^{3/2}/(3/2)+C\\
 & = & \frac{1}{3}(x^{2}+4)^{3/2}+C
\end{eqnarray*}
That \( x \) outside the square root was essential for this method to work,
since I needed to be able to replace the \( dx \) in terms of \( u \) and
\( du \), but \( du \), in terms of \( x \), has that extra \( x \). 

At first, I recommend you do all substitution integrals this way: Find what
you want \( u \) to be (looking for the \( du \) to be there (except for constants)),
then find \( du \) in terms of \( x \) and \( dx \). Then substitute the
\( u \) for the inside stuff, and replace the \( dx \) in terms of \( du \)
and \( x \). If your substitution was going to work, the \( x \) dependencies
will all vanish. Then, integrate the resulting formula in terms of \( u \)
as a variable, and finally substitute back. As you get more comfortable with
the method, though, you will easily group together the terms that need to be
there to be \( du \), up to the constants, which are easy to deal with. There
is one caveat with that, however, in that it is easy to {}``correct for{}''
the constants incorrectly. Solving for \( dx \) as above \emph{automatically}
takes care of the constants.
\end{document}
