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\centerline{Mathematics 134 -- Calculus 2 With Fundamentals}
\centerline{Solutions for Review Problems Exam 3 -- Review Sheet}
\centerline{April 5, 2018}

\vskip 10pt
\noindent
I.  Compute each of the integrals below the appropriate method or
 combination of methods.  You must show all work
for full credit. 
\begin{enumerate}
\item[A)]  Use the trigonometric reduction formulas to find
$$\int \sin^3(x) \cos^4(x)\ dx$$
\item[] {\it Solution:}  Use the SC3 formula first to get
$$\int \sin^3(x) \cos^4(x)\ dx = \frac{-\sin^2(x) \cos^5(x)}{7} + \frac{2}{7} \int \sin(x)\cos^4(x)\ dx$$
The integral that is left is $-\int u^4\ du$ for $u = \cos(x)$.  So the final answer is
$$-\frac{\sin^2(x) \cos^5(x)}{7} - \frac{2}{35}\cos^5(x) + C.$$
\item[B)]  Use the trigonometric reduction formulas to find 
$$\int \sec^5(x)\ dx$$
\item[] {\it Solution:}  Apply the ST3 formula twice, then the ST4 formula.  Answer:
$$\frac{\sec^3(x)\tan(x)}{4} + \frac{3\sec(x)\tan(x)}{8} + \frac{3}{8} \ln|\sec(x) + \tan(x)| + C$$.  
\item[C)]  $$\int \sqrt{25 - x^2}\ dx$$
\item[] {\it Solution:}  From the $25 - x^2$, we let $x = 5\sin\theta$, so $dx = 5\cos\theta\ d\theta$.  
The integral becomes
$$25 \int\cos^2(\theta)\ d\theta.$$
Then using the SC2 reduction formula
$$= 25 \left(\frac{\cos\theta\sin\theta}{2} + \frac{\theta}{2}\right) + C$$
Finally, from the substitution equation and the reference triangle, $\sin\theta = x/5$, 
$\theta = \sin^{-1}(x/5)$ and $\cos\theta = \frac{\sqrt{25 - x^2}}{5}$.  The integral 
is 
$$\frac{x\sqrt{25 - x^2}}{2} + \frac{25 \sin^{-1}(x/5)}{2} + C.$$
\item[D)]  What changes in part C if the integral is 
$$\int \sqrt{25 + x^2}\ dx$$ 
Work this one out too, start to finish.  
\item[] {\it Solution:}  The first thing that changes is that the $+$ sign indicates
we want the tangent substitution:  $x = 5\tan\theta$, so $dx = 5\sec^2\theta\ d\theta$.
The integral becomes
$$25\int \sec^3\theta\ d\theta$$
We apply the ST 3 and ST 4 reduction formulas:
$$ = \frac{25\sec\theta\tan\theta}{2} + \frac{25\ln|\sec\theta + \tan\theta|}{2} + C.$$
Finally converting back to functions of $x$, $\tan\theta = x/5$ and $\sec\theta = \sqrt{25 + x^2}{5}$, 
so the integral
$$ = \frac{x\sqrt{x^2 + 25}}{2} + \frac{25}{2}\ln\left|\frac{\sqrt{x^2 + 25}}{5} + \frac{x}{5}\right|.$$  
\item[E)]  $$\int \frac{x^3 + 2x + 1}{x^2 + 6x - 7}\ dx$$
\item[] {\it Solution:}  To apply partial fractions we need to divide first; we find 
$$\frac{x^3 + 2x + 1} = x - 6 + \frac{45 x - 41}{x^2 + 6x - 7}$$
We apply the partial fraction decomposition to the last term on the right:
$$\frac{45 x - 41}{x^2 + 6x - 7} = \frac{A}{x + 7} + \frac{B}{x - 1}$$
Hence
$$45x - 41 = A(x - 1) + B(x + 7)$$
Setting $x = -7$, we get $A = \frac{356}{8} = \frac{89}{2}$.  Similarly $x = 1$ gives $4 = 8B$, so $B = \frac{1}{2}$.
The integral is
$$\int x - 6 + \frac{315/8}{x + 7} + \frac{1/2}{x - 1}\ dx = \frac{x^2}{2} - 6x + \frac{89}{2}\ln|x + 7| + \frac{1}{2} \ln|x-1| + C.$$
\item[F)]  $$\int \frac{1}{x^2(x^2 + 6x + 10)}\ dx$$
\item[] {\it Solution:}  No division necessary here; we go immediately to the partial fractions:
$$\frac{1}{x^2 (x^2 + 6x + 10)} = \frac{A}{x} + \frac{B}{x^2} + \frac{Cx + D}{x^2+6x + 10}.$$
Clearing denominators:
\begin{align*}
1 &= Ax(x^2 + 6x + 10) + B(x^2 + 6x + 10) + (Cx + D)(x^2)\\
&= (A + C)x^3 + (6A + B + D)x^2 + (10A + 6B) x + 10B
\end{align*}
Matching coefficients shows $10B = 1$, so $B = \frac{1}{10}$.  Then from
the coefficient of $x$,  $10 A + \frac{6}{10} = 0$, so $A = \frac{-6}{100} = \frac{-3}{50}$.  From 
the coefficient of $x^3$, $A + C = 0$, so $C = \frac{3}{50}$.  Finally, from 
the coefficient of $x^2$, $\frac{-18}{50} + \frac{5}{50} + D = 0$.  So $D = \frac{13}{50}$  This gives
$$\frac{1}{x^2(x^2 + 6x + 10)} = \frac{-3/50}{x} + \frac{1/10}{x^2} + 
\frac{(3/50)x + (13/50)}{x^2 + 6x + 10}$$
We have to be slightly tricky to integrate the last term:  Complete the square on the bottom:
$x^2 + 6x + 10 = (x + 3)^2 + 1$.  We let $u = x+3$, $du = dx$, and then $x = u - 3$.  So we have
$$\frac{3}{50} \frac{u}{u^2 + 1} + \frac{4}{50} \frac{1}{u^2 + 1}$$
The first term here gives a logarithm, the second one is an inverse tangent.  The final answer
is
$$\frac{-3}{50} \ln|x| - \frac{1}{10x} + \frac{3}{100}\ln|(x+3)^2 + 1| + \frac{2}{25} \tan^{-1}(x+3) + C.$$
(Comment:  This would be a very challenging exam problem because of the trickiness of the 
arithmetic of the coefficients and the maneuvers required with completing the 
square in $x^2 + 6x + 10$.  It's good practice to work out a few this complicated, though, 
even if you are not going to see one on the exam ;)
\end{enumerate}
II.  (Improper integrals)
\begin{enumerate}
\item[A)]  Why is the following integral an \emph{improper integral}.
Decide whether it converges by setting up and evaluating the appropriate
limit(s):
$$\int_1^3 \frac{1}{(x - 3)^{2/3}}\ dx$$
\item[] {\it Solution:}  This is improper because the function has an
infinite discontinuity at $b = 3$ at the right endpoint of the interval
of integration.  We need to determine 
\begin{align*}
\lim_{b \to 3^-} \int_1^b (x-3)^{-2/3}\ dx &= 
\lim_{b \to 3^-} \left. 3(x-3)^{1/3}\right|_1^b\\
&= \lim_{b \to 3^-} 3(b - 3)^{1/3} + 3\cdot 2^{1/3}\\
&= 0 + 3\cdot 2^{1/3}.
\end{align*}
Since the limit exists, the improper integral \emph{converges}.
\item[B)]  Follow-up to part A:  For which exponents $\alpha > 0$ will the improper
$$\int_1^3 \frac{1}{(x - 3)^{2/3}}\ dx$$
\item[] {\it Solution:}  Thinking about what happened here, the limit
of the integral will exist as long as the power $1 + \alpha$ in 
the antiderivative $\frac{1}{1 + \alpha} (x - 3)^{1 + \alpha}$
is \emph{strictly positive}.  This means $0 < \alpha < 1$.  The exponent $\alpha = 1$
does not ``work'' since then the integral contains $\ln|x - 3|$ which has no limit
as $x \to 3^-$.  
\item[C)]  Why is the following integral an improper integral?
Decide whether it converges by setting up and evaluating the 
appropriate limit(s):
$$\int_0^\infty x e^{-x}\ dx.$$
\item[] {\it Solution:}  It's improper because of the infinite upper limit of 
integration.  To decide whether this one converges, we look at 
\begin{align*}
\lim_{b \to \infty} \int_0^b x e^{-x}\ dx
&=  \lim_{b \to \infty} \left. -x e^{-x} - e^{-x}\right|_0^b\quad (integrate\ by\ parts)\\
&= \lim_{b\to\infty} \frac{-b}{e^b} - \frac{1}{e^b} + 1\\
&= 0 + 0 + 1.
\end{align*}
(the first term goes to zero as $b \to \infty$ by L'Hopital's Rule).  This integral
also converges.  
\end{enumerate}
\vskip 10pt 
\noindent
III.  
\begin{enumerate}
\item[(A)]  Verify that $\int \csc\theta\ d\theta = \ln|\csc\theta - \cot\theta| + C$
by differentiating.
\item[] {\it Solution:}  By the Chain Rule and the derivative formulas 
for $\csc$ and $\cot$, 
\begin{align*}
\frac{d}{dx}\ln|\csc\theta - \cot\theta| &= 
\frac{1}{\csc\theta - \cot\theta}\cdot(-\csc\theta\cot\theta + \csc^2\theta)\\
 &= \frac{(\csc\theta - \cot\theta)\cdot \csc\theta}{\csc\theta - \cot\theta}\\
&=\csc\theta.
\end{align*}
\item[(B)] Which trigonometric substitution would you apply
to compute $\int \frac{1}{u \sqrt{a^2 - u^2}}\ du$?
What trigonometric integral do you get after making the substitution?
Complete the computation of the integral.  
\item[] {\it Solution:}  This calls for the substitution $u = a\sin\theta$, so $du = a\sin\theta\ d\theta$.
The integral becomes
$$\int \frac{a \cos\theta\ d\theta}{a \sin\theta \cdot a \cos\theta} = 
\frac{1}{a} \int \csc\theta\ d\theta.$$
Applying the formula from part (A):
$$= \frac{1}{a} \ln|\csc\theta - \cot\theta| + C.$$
The reference triangle from the substitution has $\sin\theta = u/a$, so we put $u$ on the opposite
side and $a$ on the hypotenuse.  The adjacent side is $\sqrt{a^2 - u^2}$, so 
$$\csc\theta = \frac{a}{u}\qquad {\rm and}\qquad \cot\theta = \frac{\sqrt{a^2 - u^2}}{u}.$$
The integral is 
$$\frac{1}{a} \ln\left|\frac{a}{u} - \frac{\sqrt{a^2 - u^2}}{u}\right| + C$$
\item[(C)]  Our textbook's table of integrals gives this one as
$$\frac{-1}{a}\ln\left|\frac{a + \sqrt{a^2 - u^2}}{u}\right|$$
Show that the form you got in part B is equivalent to this.  
\item[] {\it Solution:}  By properties of logarithms:
$$\frac{-1}{a}\ln\left|\frac{a + \sqrt{a^2 - u^2}}{u}\right| = 
\frac{1}{a}\ln\left|\frac{u}{a + \sqrt{a^2 - u^2}}\right|$$
Inside the logarithm, multiply top and bottom of the fraction by the conjugate radical
 $a - \sqrt{a^2 - u^2}$.
On the bottom, expand to $(a^2 - (a^2 - u^2)) = u^2$.  Then we can cancel one power
of $u$ to yield
$$\frac{1}{a}\ln\left|\frac{a - \sqrt{a^2 - u^2}}{u}\right|$$
Separating the inside into two fractions with denominator $u$ gives the form we derived in part (B).
\end{enumerate}
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