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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{The definition of the Riemann Integral}

\begin{example}
Show that \[
\int _{0}^{2\pi }\sin ^{2}xdx=\pi .\]
 The same is true for \( \cos ^{2}x \), \[
\int _{0}^{2\pi }\cos ^{2}xdx=\pi .\]

\end{example}

\subsubsection*{Solution}

This is a classic integral. It is not only one of the few trigonometric integrals
that can be solved with absolutely no techniques of integration, but it is one
of the most useful. In later courses, you will run across these two integrals,
and simple variations of them, over and over again. To be able to evaluate those
integrals on sight saves a lot of headaches.

There is a way to solve these integrals using techniques of integration, but
that is more than you need to do here. This method specifically will work for
integrals over a full period of the function, not for more general intervals.

First, graph \( y=\sin ^{2}(x) \) over that interval. 

\vspace{0.3cm}
{\par\centering \includegraphics{sine-squared.ps} \par}
\vspace{0.3cm}

Note that this graph is basically two {}``hills{}''. Both are above the axis
since \( \sin ^{2}(x) \) is of course always positive. Now compare that graph
to that of \( \cos ^{2}(x) \) over the same interval, 

\vspace{0.3cm}
{\par\centering \includegraphics{cosine-squared.ps} \par}
\vspace{0.3cm}

Notice that the different-colored areas of this graph can be put into correspondence
with those of \( \sin ^{2}(x) \), so that \[
\int _{0}^{2\pi }\sin ^{2}(x)dx=\int _{0}^{2\pi }\cos ^{2}(x)dx.\]
On the other hand, since \[
\sin ^{2}(x)+\cos ^{2}(x)=1\]
 at each point (\emph{the} basic trigonometric identity), \begin{eqnarray*}
\int _{0}^{2\pi }\sin ^{2}(x)dx+\int _{0}^{2\pi }\cos ^{2}(x)dx & = & \int _{0}^{2\pi }\left( \sin ^{2}(x)+\cos ^{2}(x)\right) dx\\
 & = & \int _{0}^{2\pi }1\, dx\\
 & = & 2\pi .
\end{eqnarray*}
Since the two integrals are the same, and the sum of the two comes up to \( 2\pi  \),
each one must be half that, \( \pi  \), that is, \[
\int _{0}^{2\pi }\sin ^{2}(x)dx=\pi =\int _{0}^{2\pi }\cos ^{2}(x)dx.\]
There are several similar integrals to these, the most general being:

\[
\int _{a}^{a+n\pi }\sin ^{2}(kx)dx=\frac{n\pi }{2}=\int _{a}^{a+n\pi }\cos ^{2}(kx)dx,\]
where \( a \) is any number, but \( n \) and \( k \) have to be integers,
with \( n \) positive. The proof is exactly the same, but you do have to be
a bit careful to make sure you have the right pieces to make it work.
\end{document}
