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\title{Math 1a.  Lecture 4 \\
Additional Techniques of Integration \\
(Trigonometric Integrals and Trigonometric Substitution)}

\author{T. Judson}

\date{Spring 2006}


\begin{document}

\maketitle


\section{Goals}

\begin{itemize}

\item
To be able to evaluate trigonometric integrals.

\item
To be able to evaluate definite and indefinite integrals using trigonometric substitution.


\end{itemize} 



\section{Trigonometric Integrals}


\begin{itemize}

\item
If the integrand contains an odd power, it can usually be reduced to a substitution problem if we use the identity, $\cos^2 x + \sin^2 x = 1$. For example,
$$
\int \sin^3 x \, dx = \int (1 - \cos^2 x) \sin x \, dx
$$

\item
If the integrand contains only even powers such as
$$
\int \cos^2 x \, dx,
$$
we can use the half-angle formulas,
\begin{eqnarray*}
\sin^2 x & = & \frac{1}{2}(1 - \cos 2x) \\
\cos^2 x & = & \frac{1}{2}(1 + \cos 2x).
\end{eqnarray*}
These follow from the double angle formula for $\cos x$.


\end{itemize}





\section{ Trigonometric Substitution}


Integrals containing an expression of the form $\sqrt{a^2 - x^2}$,  $\sqrt{x^2 - a^2}$, or $\sqrt{a^2 + x^2}$ can be evaluated using trigonometric substitution.\footnote{Stewart's explanation is lacking here.}  Consider the following example,
$$
\int \frac{x^3}{\sqrt{1 - x^2}} \, dx,
$$
where $|x| < 1$.
We can make the substitution
\begin{eqnarray*}
x & = & \sin \theta \\
dx & = & \cos \theta \, d\theta.
\end{eqnarray*}
Then
\begin{eqnarray*}
\int \frac{x^3}{\sqrt{1 - x^2}} \, dx
& = & 
\int \frac{\sin^3 \theta \cos \theta}{\sqrt{1 - \sin^2 \theta}} \, d\theta \\
& = &
\int \sin^3 \theta \, d\theta \\
& = &
\frac{1}{3} \cos^3 \theta - \cos \theta + C.
\end{eqnarray*}
Since $\cos \theta = \sqrt{1 - x^2}$,
$$
\int \frac{x^3}{\sqrt{1 - x^2}} \, dx
=
\frac{1}{3}(1 - x^2)^{3/2} - \sqrt{1 - x^2} + C.
$$
Use the picture below.

\bigskip

\begin{center}
\includegraphics[width=3in]{Fig_4_1}
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\section{Worksheet Exercises}

Evaluate each of the following integrals
\begin{enumerate}

\item 
$\displaystyle \int \cos^3 x\sin^4 x \, dx$



\item 
$\displaystyle \int \sin^2 x \cos^2 x \, dx$



\item 
$\displaystyle \int \sec^3 x \tan^3 x \, dx$.



\item 
$\displaystyle \int \sec \theta \, d\theta =  \int \frac{\cos \theta}{\cos^2 \theta} \, d\theta =  \int \frac{\cos \theta}{1 - \sin^2 \theta} \, d\theta $.  {\em Hint}: Make use of the identity
$$
\frac{2}{1 - u^2} = \frac{1}{1 - u} + \frac{1}{1+ u}.
$$




\item 
$\displaystyle \int \sqrt{4 - x^2} \, dx$



\item 
$\displaystyle \int \frac{1}{\sqrt{4 + x^2}} \, dx$


\end{enumerate}


\section*{References}

\begin{itemize}

\item
\S 5.7 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 13, 2006}



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