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\begin{document}
\begin{center}

\textbf{College of the Holy Cross, Spring 2018} \\
\textbf{Math 134 Midterm Exam 3 -- Solutions} \\
\textbf{Friday, April 13}
\end{center}

\vskip 10pt
\noindent
I.  Compute each of the following integrals using the trigonometric reduction formulas (on separate sheet)
and other methods as appropriate.
\begin{enumerate}
\item[A.] (15) $\displaystyle\int \sin^4 (5x)\ dx$
\item[] {\it Solution:}  First let $u = 5x$ and then $dx = (1/5)\ du$.  Then using 
the SC1 reduction formula, the integral is 
$$\frac{1}{5}\left(\frac{-\sin^3(u)\cos(u)}{4} + \frac{3}{4}\left(\frac{-\sin(u)\cos(u)}{2} + \frac{u}{2}\right)\right) + C.$$
Substituting $u = 5x$ back in and simplifying the final answer is 
$$\frac{-\sin^3(5x)\cos(5x)}{20} + \frac{-3\sin(5x)\cos(5x)}{40} + \frac{3x}{8} + C$$
\item[B.] (15) $\displaystyle\int x \sec^4 (3x^2) \ dx$
\item[] {\it Solution:}  First let $u = 3x^2$, so $x\ dx = \frac{1}{6}\ du$.  Then using the 
ST3 reduction formula, the integral is
$$\frac{1}{6} \int \sec^4(u)\ du = \frac{\tan(u)\sec^2(u)}{18} + \frac{\tan(u)}{9} + C$$
This is 
$$= \frac{\tan(3x^2)\sec^2(3x^2)}{18} + \frac{1}{9}\tan(3x^2) + C.$$
\end{enumerate}
\vskip 10pt
\noindent
II.  (Trigonometric Substitutions)
\begin{enumerate}
\item[A.]  (10) What trigonometric substitution would you use to evaluate
$$\int \frac{x^2}{\sqrt{64 + x^2}}\ dx?$$
Make the substitution and simplify so there is no square root in the resulting trigonometric
integral.  
\item[] {\it Solution:}  The form of $\sqrt{64 + x^2}$ indicates that we want 
$x = 8\tan\theta$.  Then $dx = 8\sec^2\theta\ d\theta$, and the integral becomes
$$\int \frac{(8\tan\theta)^2}{\sqrt{64 + 64\tan^2\theta}}\ 8\sec^2\theta\ d\theta = 
64 \int \frac{\tan^2\theta}{\sec\theta} \sec^2\theta\ d\theta$$
This could also be simplified to 
$$64\int \tan^2\theta \sec\theta\ d\theta.$$
\item[B.]  (10) Suppose you had used the trig substitution $x = 3\sin\theta$ and you integrated
the resulting trigonometric integral to get
$$\ln |\csc\theta - \cot\theta| + C$$
What is the integral expressed as a function of $x$?
\item[] {\it Solution:}  The reference triangle has $x$ on the opposite side, $3$ on 
the hypotenuse, and $\sqrt{9 - x^2}$ on the adjacent side.  Hence $\csc\theta$ equals
the hypotenuse over the opposite side and $\cot\theta$ equals the adjacent side
over the opposite side.  The function is 
$$\ln \left|\frac{3}{x} - \frac{\sqrt{9 - x^2}}{x}\right| + C.$$
\end{enumerate}
\vskip 10pt
\noindent
III.  (Partial Fractions)
\begin{enumerate}
\item[A.] (5) To integrate $\displaystyle\frac{x^4 + 1}{x^3 + 7x + 1}$, what would 
be the first step?  Carry the step out and give the rational function to which you would apply the 
partial fraction decomposition.
\item[] {\it Solution:}  Since the degree of the top is 4 and the degree of the bottom is 3, we
need to divide $x^3 + 7x + 1$ into $x^4 + 1$.  The quotient is $x$ and the remainder is
$-7x^2 - x + 1$.  The result of the division is: 
$$ \frac{x^4 + 1}{x^3 + 7x + 1} = x + \frac{-7x^2 - x + 1}{x^3 + 7x + 1}.$$
\item[B.] (10) To decompose $\displaystyle\frac{1}{(x+4)^2(x + 1)^3(x^2 + 9)}$ into partial fractions, what would the form of the fractions be (leave coefficients as undetermined; 
\emph{do not try to solve for them}).  
\item[] {\it Solution:}  From the form of the factorization, 
$$\frac{A}{x + 4} + \frac{B}{(x + 4)^2} + \frac{C}{x+1} + \frac{D}{(x+1)^2} + \frac{E}{(x + 1)^3} 
+ \frac{Fx + G}{x^2 + 9}.$$
\item[C.]  (10)  Determine the values $A,B,C$ making 
$$\frac{3x + 1}{x(x^2 + 4)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 4}$$
\item[] {\it Solution:}  Clearing denominators, we have 
\begin{align*} 
3x + 1 &= A(x^2 + 4) + (B x + C)x\\
&= (A + B)x^2 + C x + 4A.
\end{align*}
Equating coefficients of like powers of $x$ gives:
$$A + B = 0, C = 3, 4A = 1.$$
The second and third equations give $A = 1/4$ and $C = 3$.  Then from 
the first equation $B = -1/4$.

\item[D.]  (15)  You have decomposed a rational function $f(x)$ into partial fractions as
$$f(x) = x^2 + 3x + \frac{1}{(x+3)^2} + \frac{2}{(x+3)^3}+ \frac{4x + 3}{x^2 + 1}.$$
What is $\int f(x)\ dx$?
\item[] {\it Solution:}  By the power rule ($u$-substitution) and the inverse tangent form,
$$\int f(x)\ dx = \frac{x^3}{3} + \frac{3x^2}{2} - \frac{1}{x+3} - \frac{1}{(x+3)^2} + 
2\ln|x^2 + 1| + 3\tan^{-1}(x) + C$$
\end{enumerate}
\vskip 10pt
\noindent
IV.  (Improper Integrals)  
\begin{enumerate}
\item[A.] (5)  Is there a finite area between the $x$-axis and the graph 
$y = \frac{1}{x^{2/3}}$ for $0 < x \le 1$?  Set up the appropriate (improper)
integral and determine if it converges.
\item[] {\it Solution:}  The necessary integral is 
$$\int_0^1 \frac{1}{x^{2/3}}\ dx = \int_0^1 x^{-2/3}\ dx$$
This is improper because of the discontinuity at the lower limit of integration.  So
we must determine whether
$$\lim_{a\to 0^+} \int_a^1 x^{-2/3}\ dx$$ 
exists.  Computing the integral we have 
$$\lim_{a\to 0^+} \left.3x^{1/3}\right|_a^1 = \lim_{a\to 0^+} 3 - 3 a^{1/3} = 3$$
Since the limit exists, the integral converges.  The area is finite and equals 3.
\item[B.] (5)  Does the solid of revolution obtained by rotating the region from part A
about the $x$-axis have a finite volume or not?  Why?
\item[] {\it Solution:}  The volume would be computed by 
$$\int_0^1 \pi (x^{-2/3})^2\ dx = \pi \int_0^1 x^{-4/3}\ dx.$$
This is improper for the same reason as the integral in part A and we determine whether
it converges by a similar method.  But
$$\int_0^1 x^{-4/3}\ dx = \lim_{a\to 0^+} \left. -3x^{-1/3}\right|_a^1 = \lim_{a\to 0^+} -3 + 3 a^{-1/3},$$
which does not exist because the $a^{-1/3} = \frac{1}{a^{1/3}}$ is now in the denominator.
This integral \emph{diverges}(!)  In this case, we get a finite area such that the corresponding
solid of revolution does not have finite volume.
\end{enumerate}
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