\documentclass[12pt]{article}
\usepackage{graphicx}
\usepackage{amsmath}

\begin{document}
\centerline{MATH 134 -- Calculus with Fundamentals 2}
\centerline{Discussion Day on Trigonometric Substitution Integrals}
\centerline{March 21, 2018}
\vskip 10pt
\noindent
{\it Background}
\vskip 10pt
Recall that we saw some first examples of the \emph{trigonometric substitution} 
method last time.  The basic outline of this method is that for
integrals involving
\begin{itemize}
\item $\sqrt{a^2 - x^2}$, we let $x = a\sin\theta$ and $dx = a\cos\theta\ d\theta$
\item $\sqrt{a^2 + x^2}$, we let $x = a\tan\theta$ and $dx = a\sec^2\theta\ d\theta$
\item $\sqrt{x^2 - a^2}$, we let $x = a\sec\theta$ and $dx = a\sec\theta\tan\theta\ d\theta$
\end{itemize}
(The last one is only included for completeness and possible future reference. We will
only be doing integrals involving the $\sin$ and $\tan$ subsitutions.)
We simplify, then apply our trig reduction formulas from last week.  The last step is
to convert back to functions of $x$ using the reference triangle corresponding to the 
substition used:
\begin{itemize}
\item For the $x = a\sin\theta$ substitution, put $x$ on opposite side, $a$ on hypotenuse, then $(adj) = 
\sqrt{a^2 - x^2}$, so you can read off any trig function of $\theta$ from the triangle and $\theta = \sin^{-1}(x/a)$
\item For the $x = a\tan\theta$ substitution, put $x$ on opposite side, $a$ on adjacent, then $(hyp) = \sqrt{x^2 + a^2}$, 
so you can read off any trig function of $\theta$ from the triangle and $\theta = \tan^{-1}(x/a)$
\item For the $x = a\sec\theta$ substitution, put $x$ on hypotenuse, $a$ on adjacent, then $(opp) = \sqrt{x^2 - a^2}$, 
so you can read off any trig function of $\theta$ from the triangle and $\theta = \sec^{-1}(x/a)$.  (This case is included
for completeness, you are only responsible for the first two.) 
\end{itemize}

\vskip 10pt 
\noindent
{\it Questions}
\begin{enumerate}
\item[(A)]  Using the appropriate trigonometric substitution compute
$$\int \frac{1}{\sqrt{x^2 + 25}}\ dx$$
\item[(B)]  Using the appropriate trigonometric substitution compute
$$\int \frac{1}{(36 - x^2)^{3/2}}\ dx$$
\item[(C)]  You want to divide a circular pizza with radius 9in (say with
outer crust along the circle $x^2 + y^2 = 81$)  into
three exactly equal pieces with a knife or pizza cutter.  
One way, of course, is to divide it into
three sectors each with angle exactly $2\pi/3$ by cutting from the 
center to the outer edge.  But another (easier?) way would 
be to cut the pizza with two parallel strokes, say at $x = \pm a$
so that the three strips $-9 \le x \le -a$, $-a \le x \le a$, and $a \le x \le 9$
all have the same area.  Set up an integral with $a$ in the limits of integration, 
evaluate it using an appropriate trigonometric substitution, and find
an equation to solve for the $a$ giving three strips of exactly equal area.
Find an approximation for $a$ by using a graphing calculator.  
\end{enumerate}
\end{document}

\begin{figure}
\begin{center}
\includegraphics[width=.8\linewidth]{needleregion.png}
\end{center}
\caption{The region for problem 4.}
\end{figure}

\end{document}



