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\centerline{MATH 136 -- Calculus 2}
\centerline{Practice Day on Trigonometric Substitution Integrals}
\centerline{February 14, 2020}
\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}
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 $\theta = \sec^{-1}(x/a)$.  
\end{itemize}

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\noindent
{\it Questions}
\begin{enumerate}
\item[(A)]  Using the appropriate trigonometric substitution compute
$$\int \frac{1}{\sqrt{x^2 + 25}}\ dx$$
\item[]  {\it Answer:}  Let $x = 5\tan(\theta)$ and $dx = 5\sec^2(\theta)\ d\theta$.
The integral becomes
\begin{align*}
\int \frac{5\sec^2(\theta)\ d\theta)}{5\sec(\theta)} &= \int\sec(\theta)\ d\theta\\
&= \ln|\sec(\theta) + \tan(\theta)| + C\\
&= \ln|\sqrt{x^2 + 25} + x| + C
\end{align*}
Note: You could also write this as 
$$\ln\left|\frac{\sqrt{x^2 + 25}}{5} + \frac{x}{5}\right| + C$$
but by properties of logarithms, the $-\ln(5)$ from the denominator can
be absorbed into the constant of integration, so either form is correct.
\item[(B)]  Using the appropriate trigonometric substitution compute
$$\int \frac{1}{(36 - x^2)^{3/2}}\ dx$$
\item[] {\it Answer:}  Let $x = 6\sin(\theta)$ and $dx = 6\cos(\theta)\ d\theta$. 
The integral becomes 
$$\frac{1}{36} \int \cos^{-2}(\theta)\ d\theta = \frac{1}{36}\int \sec^2(\theta)\ d\theta = 
\frac{1}{36} \tan(\theta) + C.$$
After setting up the triangle, we see $\sin(\theta) = \frac{x}{6}$ is the opposite over 
hypotenuse.  The adjacent side is $\sqrt{36 - x^2}$ and the tangent is 
opposite over adjacent or 
$$ = \frac{x}{36 \sqrt{36 - x^2}} + C.$$  
\item[(C)]  Using the appropriate trigonometric substitution compute
$$\int \frac{1}{\sqrt{x^2 - 16}}\ dx$$
\item[] {\it Answer:}  The form $x^2 - 16$ under the square root calls
for the substitution $x = 4\sec(\theta)$, so $dx = 4\sec(\theta)\tan(\theta)\ d\theta$.
The integral becomes
$$\int \frac{4\sec(\theta)\tan(\theta)\ d\theta}{4\tan(\theta)} = \int \sec(\theta)\ d\theta
= \ln|\sec(\theta) + \tan(\theta)| + C.$$
To convert back to functions of $x$, from $\sec(\theta) = \frac{x}{4}$, we set up 
a triangle containing the angle $\theta$ with hypotenuse $x$ and adjacent side $4$.
The opposite side is $\sqrt{x^2 - 16}$.  
Then $\sec(\theta) = \frac{x}{4}$ and $\tan(\theta) = \frac{\sqrt{x^2 - 16}}{4}$.  As
in the integral from (A), the denominators can be absorbed into the constant 
of integration and the answer can be written as 
$$= \ln|x + \sqrt{x^2 - 16}| + C$$
\item[(D)]  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.  
\item[] {\it Answer:}  By the circular symmetry, we only need to find $a$ to 
make: 
$$\int_{-a}^a \sqrt{81 - x^2}\ dx = \int_a^9 \sqrt{81 - x^2}\ dx$$
The integrals can be computed using the FTC, part I and the antiderivative formula
$$\int \sqrt{81 - x^2}\ dx = \frac{x\sqrt{81 - x^2}}{2} + \frac{81\sin^{-1}(x/9)}{2}$$
(from the trig substition $x = 9\sin(\theta)$).  We get an equation:
$$\frac{3a}{2}\sqrt {81 - a^2}+{\frac {243\,\sin^{-1} \left( a/9 \right) }{2}
}-{\frac {81\,\pi}{4}} = 0$$
If you plot this function of $a$, you can see that the curve crosses the horizonal
axis at about $a \doteq 2.384$.  Slicing the pie at $x = \pm a = \pm 2.384$ will yield
three pieces of equal areas (but different shapes of course!)

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

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