\documentclass[10pt,letter]{exam}
\usepackage{amsmath}
\usepackage{tabularx,ragged2e}
\usepackage{hyperref}
\usepackage{color}
\begin{document}

\begin{coverpages}
\begin{center}\huge MATH 3315 - EXAM 1 SPRING 2017\\
SOLUTION
\end{center}

\begin{flushleft}\vspace{0.5in}
Friday 10 February 2017 \\
Instructor: Tom Cuchta
\end{flushleft}

\textbf{Instructions:}
\begin{itemize}
	\item  Show all work, clearly and in order, if you want to get full
	credit. If you claim something is true \textbf{you must show work backing up your claim}. I reserve the right to take off points if I cannot see how you arrived at your answer (even if your final answer is correct).
	\item Justify your answers algebraically whenever possible to ensure full credit. 	
	\item  Circle or otherwise indicate your final answers.
	\item  Please keep your written answers brief; be clear and to the point.
	\item  Good luck!
\end{itemize}
\vspace*{10pt}


\end{coverpages}

\begin{questions}

\question[30] Compute $\ldots$
\begin{parts}
\part[4] $\dfrac{\mathrm{d}}{\mathrm{d}x} \ln(x^2+3x+2)$ \\
\textit{Solution:} \textbf{This problem resembles \texttt{\#}41 and \texttt{\#}55 in {\color{blue} \href{http://tomcuchta.com/teach/classes/2017/MATH-3315-Spring-2017-FairmontState/homework/hw1.php}{HW1}}}.\\
{\color{blue} \href{http://www.wolframalpha.com/input/?i=d%2Fdx+%5Bln(x%5E2%2B3x%2B2)%5D}{Calculate}} 
$$\begin{array}{ll}
\dfrac{\mathrm{d}}{\mathrm{d}x} \ln(x^2+3x+2) &= \dfrac{1}{x^2+3x+2} \dfrac{\mathrm{d}}{\mathrm{d}x} (x^2+3x+2) \\
&= \dfrac{2x+3}{x^2+3x+2}.
\end{array}$$
\part[4] $\dfrac{\mathrm{d}}{\mathrm{d}x} (e^{2x}+e^{-3x})^2$\\
\textit{Solution:}  \textbf{This problem resembles \texttt{\#}43 and \texttt{\#}46 in {\color{blue} \href{http://tomcuchta.com/teach/classes/2017/MATH-3315-Spring-2017-FairmontState/homework/hw2.php}{HW2}}}.\\
{\color{blue} \href{http://www.wolframalpha.com/input/?i=d%2Fdx+%5B+(e%5E(2x)%2Be%5E(-3x))%5E2+%5D}{Calculate}}
$$\begin{array}{ll}
\dfrac{\mathrm{d}}{\mathrm{d}x} (e^{2x}+e^{-3x})^2 &=  2(e^{2x}+e^{-3x}) \dfrac{\mathrm{d}}{\mathrm{d}x} [e^{2x}+e^{-3x}] \\
&=2 (e^{2x}+e^{-3x}) (2e^{2x}-3e^{-3x}). \\
\end{array}$$
\textit{(note: the solution at the link, $2e^{-6x}(e^{6x}+1)(2e^{5x}-3),$ is the same as this -- to see it multiply three of the $e^{-x}$ factors in $e^{-6x}$ into each of the two other factors respectively...same result!)}
\part[4] $\dfrac{\mathrm{d}}{\mathrm{d}x} \arctan(\ln(x))$ \\
\textit{Solution:} \textbf{This problem resembles \texttt{\#}39 and \texttt{\#}43 in {\color{blue} \href{http://tomcuchta.com/teach/classes/2017/MATH-3315-Spring-2017-FairmontState/homework/hw3.php}{HW3}}}.\\
{\color{blue}\href{http://www.wolframalpha.com/input/?i=d%2Fdx+[+arctan(ln(x))+]}{Calculate}}
$$\begin{array}{ll}
\dfrac{\mathrm{d}}{\mathrm{d}x} \arctan(\ln(x)) &= \dfrac{1}{1+(\ln(x))^2} \dfrac{\mathrm{d}}{\mathrm{d}x} \ln(x) \\
&=\dfrac{1}{x(1+(\ln(x)))^2}.
\end{array}$$
\part[4] $\dfrac{\mathrm{d}}{\mathrm{d}x} 7^{3x-8}$ \\
\textit{Solution:} \textbf{This problem resembles \texttt{\#}40 in {\color{blue} \href{http://tomcuchta.com/teach/classes/2017/MATH-3315-Spring-2017-FairmontState/homework/hw2.php}{HW2}}}.\\
Rewrite
$$7^{3x-8} = e^{\ln(7^{3x-8})}=e^{(3x-8)\ln(7)}.$$
Now compute the {\color{blue} \href{http://www.wolframalpha.com/input/?i=d%2Fdx+%5B7%5E(3x-8)%5D}{derivative}}
$$\begin{array}{ll}
\dfrac{\mathrm{d}}{\mathrm{d}x} 7^{3x-8} &= \dfrac{\mathrm{d}}{\mathrm{d}x} e^{(3x-8)\ln(7)} \\ 
&= e^{(3x-8)\ln(7)} \dfrac{\mathrm{d}}{\mathrm{d}x} \left[ (3x-8)\ln(7) \right] \\ 
&=3 \ln(7) 7^{3x-8}.
\end{array}$$

\part[4] $\dfrac{\mathrm{d}}{\mathrm{d}x} \left[ \cosh(x)\sinh(x) \right]$ \\
\textit{Solution:} \textbf{This problem resembles \texttt{\#}30 in {\color{blue} \href{http://tomcuchta.com/teach/classes/2017/MATH-3315-Spring-2017-FairmontState/homework/hw3.php}{HW3}}}.\\
{\color{blue} \href{http://www.wolframalpha.com/input/?i=d%2Fdx+%5Bcosh(x)sinh(x)%5D}{Calculate}} 
$$\begin{array}{ll}
\dfrac{\mathrm{d}}{\mathrm{d}x} \cosh(x)\sinh(x) &= \dfrac{\mathrm{d}}{\mathrm{d}x} [\cosh(x)] \sinh(x) + \cosh(x) \dfrac{\mathrm{d}}{\mathrm{d}x} [\sinh(x)] \\
&=\sinh^2(x) + \cosh^2(x).
\end{array}$$
\textit{note: the link claims the answer is $\cosh(2x)$. This can be checked by algebraically expanding the expression $\sinh^2(x)+\cosh^2(x)$ with the definitions $\sinh(x)=\dfrac{e^{x}-e^{-x}}{2}$ and $\cosh(x) = \dfrac{e^x + e^{-x}}{2}$.}
\part[5] $\dfrac{\mathrm{d}}{\mathrm{d}x} \left[ e^x \mathrm{arcsinh}(x^2) \right]$ \\
\textit{Solution:} \textbf{This problem resembles \texttt{\#}70 in {\color{blue} \href{http://tomcuchta.com/teach/classes/2017/MATH-3315-Spring-2017-FairmontState/homework/hw3.php}{HW3}}}.\\
{\color{blue} \href{http://www.wolframalpha.com/input/?i=d%2Fdx+%5Be%5Ex+arcsinh(x%5E2)+%5D}{Calculate}} 
$$\begin{array}{ll}
\dfrac{\mathrm{d}}{\mathrm{d}x} e^x \mathrm{arcsinh}(x^2) &= \dfrac{\mathrm{d}}{\mathrm{d}x}[e^x]\mathrm{arcsinh}(x^2) + e^x \dfrac{\mathrm{d}}{\mathrm{d}x} [\mathrm{arcsinh}(x^2)] \\
&=e^x \mathrm{arcsinh(x^2)} + \dfrac{e^x}{\sqrt{1+(x^2)^2}} \dfrac{\mathrm{d}}{\mathrm{d}x} [x^2] \\
&=e^x \mathrm{arcsinh(x^2)} + \dfrac{2x e^x}{\sqrt{1+x^4}}. \\
\end{array}$$
\part[5] $\dfrac{\mathrm{d}}{\mathrm{d}x} \ln(\ln(\ln(x)))$ \\
\textit{Solution:}  \textbf{This problem resembles \texttt{\#}41 and \texttt{\#}55 in {\color{blue} \href{http://tomcuchta.com/teach/classes/2017/MATH-3315-Spring-2017-FairmontState/homework/hw1.php}{HW1}}}.\\ {\color{blue} \href{http://www.wolframalpha.com/input/?i=d%2Fdx%5Bln(ln(ln(x)))%5D}{Calculate}}
$$\begin{array}{ll}
\dfrac{\mathrm{d}}{\mathrm{d}x} \ln(\ln(\ln(x))) &= \dfrac{1}{\ln(\ln(x))} \dfrac{\mathrm{d}}{\mathrm{d}x} [\ln(\ln(x))] \\
&= \dfrac{1}{\ln(\ln(x))\ln(x)} \dfrac{\mathrm{d}}{\mathrm{d}x} \ln(x) \\
&=\dfrac{1}{x\ln(\ln(x))\ln(x)}.
\end{array}$$
\end{parts}

\question[35] Compute the following indefinite integrals.
\begin{parts}
\part[7] $\displaystyle\int \dfrac{1}{1+9x^2} \mathrm{d}x$\\
\textit{Solution:} \textbf{This problem resembles \texttt{\#}23 and \texttt{\#}86  in {\color{blue} \href{http://tomcuchta.com/teach/classes/2017/MATH-3315-Spring-2017-FairmontState/homework/hw3.php}{HW3}}}.\\
Notice that $9x^2=(3x)^2$. This means the integrand is 
$$\dfrac{1}{1+9x^2} = \dfrac{1}{1+(3x)^2}.$$
So we will do a $u$-substitution by letting $u=3x$ so that $\mathrm{d}u=3 \mathrm{d}x$, or in other words, $\dfrac{1}{3} \mathrm{d}u = \mathrm{d}x$. Now {\color{blue} \href{http://www.wolframalpha.com/input/?i=integral+of+1%2F(1%2B9x%5E2)+dx}{calculate}}
$$\begin{array}{ll}
\displaystyle\int \dfrac{1}{1+9x^2} \mathrm{d}x &= \displaystyle\int \dfrac{1}{1+(3x)^2} \mathrm{d}x \\
&\stackrel{u=3x}{=} \dfrac{1}{3} \displaystyle\int \dfrac{1}{1+u^2} \mathrm{d}u \\
&=\dfrac{1}{3} \arctan(u) + C \\
&\stackrel{u=3x}{=} \dfrac{1}{3} \arctan(3x)+C.
\end{array}$$
\part[7] $\displaystyle\int \dfrac{e^x}{1+e^{x}} \mathrm{d}x$\\
\textit{Solution:} \textbf{This problem resembles \texttt{\#}99 and \texttt{\#}103 in {\color{blue} \href{http://tomcuchta.com/teach/classes/2017/MATH-3315-Spring-2017-FairmontState/homework/hw2.php}{HW2}}}.\\
Let $u=1+e^x$ so that $\mathrm{d}u = e^x \mathrm{d}x$. Now {\color{blue} \href{http://www.wolframalpha.com/input/?i=integral+of+e%5Ex%2F(1%2Be%5Ex)+dx}{calculate}}
$$\begin{array}{ll}
\displaystyle\int \dfrac{e^x}{1+e^x} \mathrm{d}x &\stackrel{u=1+e^x}{=} \displaystyle\int \dfrac{1}{u} \mathrm{d}u \\
&= \ln(|u|)+C \\
&\stackrel{u=1+e^x}{=}\ln(|1+e^x|)+C \\
&= \ln(1+e^x)+C, 
\end{array}$$
where the last line is true because $e^x$ never touches the $x$-axis (i.e. it is never zero). 
\part[7] $\displaystyle\int 3^{\cos(x)}\sin(x) \mathrm{d}x$ \\
\textit{Solution:} \textbf{This problem resembles \texttt{\#}78 in {\color{blue} \href{http://tomcuchta.com/teach/classes/2017/MATH-3315-Spring-2017-FairmontState/homework/hw2.php}{HW2}}}.\\
Let $u=\cos(x)$ so that $\mathrm{d}u=-\sin(x) \mathrm{d}x$, or equivalently, $-\mathrm{d}u=\sin(x) \mathrm{d}x$.  Now {\color{blue} \href{http://www.wolframalpha.com/input/?i=integral+of+3%5E(cos(x))*sin(x)+dx}{calculate}}
$$\begin{array}{ll}
\displaystyle\int 3^{\cos(x)}\sin(x) \mathrm{d}x &\stackrel{u=\cos(x)}{=} -\displaystyle\int 3^u \mathrm{d}u \\
&= -\displaystyle\int e^{u\ln(3)} \mathrm{d}u \\
&\stackrel{w=u\ln(3)}{=} \dfrac{-1}{\ln(3)} \displaystyle\int e^w \mathrm{d}w \\
&=\dfrac{-1}{\ln(3)} e^w + C \\
&\stackrel{w=u\ln(3)}{=} -\dfrac{e^{u\ln(3)}}{\ln(3)} + C \\
&\stackrel{u=\cos(x)}{=} -\dfrac{3^{\cos(x)}}{\ln(3)} + C
\end{array}$$

\part[7] $\displaystyle\int \dfrac{5x^4+2x+1}{x^5+x^2+x} \mathrm{d}x$ \\
\textit{Solution:} \textbf{This problem resembles \texttt{\#}14 in {\color{blue} \href{http://tomcuchta.com/teach/classes/2017/MATH-3315-Spring-2017-FairmontState/homework/hw1.php}{HW1}}}.\\
Let $u=x^5+x^2+x$ so that $\mathrm{d}u=5x^4+2x+1 \mathrm{d}x$. Now {\color{blue} \href{http://www.wolframalpha.com/input/?i=integral+of+(5x%5E4%2B2x%2B1)%2F(x%5E5%2Bx%5E2%2Bx)+dx}{calculate}}
$$\begin{array}{ll}
\displaystyle\int \dfrac{5x^4+2x+1}{x^5+x^2+x} \mathrm{d}x &\stackrel{u=x^5+x^2+x}{=} \displaystyle\int \dfrac{1}{u} \mathrm{d}u \\
&= \ln(|u|) + C \\
&\stackrel{u=x^5+x^2+x}{=} \ln(|x^5+x^2+x|) + C
\end{array}$$
\textit{note: the solution at the link goes a little bit farther by factoring out an $x$ inside of the logarithm and writing
$\ln(|x| |x^4+x+1|)=\ln(|x|)+\ln(|x^4+x+1|)$}
\part[7] $\displaystyle\int \dfrac{\sinh(x)}{\cosh(x)} \mathrm{d}x$ \\
\textit{Solution:} \textbf{This problem resembles \texttt{\#}49 in {\color{blue} \href{http://tomcuchta.com/teach/classes/2017/MATH-3315-Spring-2017-FairmontState/homework/hw3.php}{HW3}}}.\\
Let $u=\cosh(x)$ so that $\mathrm{d}u=\sinh(x) \mathrm{d}x$. Now {\color{blue} \href{http://www.wolframalpha.com/input/?i=integral+of+sinh(x)%2Fcosh(x)+dx}{calculate}}
$$\begin{array}{ll}
\displaystyle\int \dfrac{\sinh(x)}{\cosh(x)} \mathrm{d}x &\stackrel{u=\cosh(x)}{=} \displaystyle\int \dfrac{1}{u} \mathrm{d}u \\
&= \ln(|u|) + C \\
&\stackrel{u=\cosh(x)}{=} \ln(|\cosh(x)|)+C \\
&= \ln(\cosh(x)) + C. 
\end{array}$$
\end{parts}


\question[35] Compute the following definite integrals.
\begin{parts}
\part[7] $\displaystyle\int_0^1 \dfrac{1}{\sqrt{36+x^2}} \mathrm{d}x$ \\
\textit{Solution:} \textbf{This problem resembles \texttt{\#}23 and \texttt{\#}86  in {\color{blue} \href{http://tomcuchta.com/teach/classes/2017/MATH-3315-Spring-2017-FairmontState/homework/hw3.php}{HW3}}}.\\
Note that $36=6^2$ and {\color{blue} \href{http://www.wolframalpha.com/input/?i=integral+from+x%3D0+to+x%3D1+of+1%2Fsqrt(36%2Bx%5E2)+dx}{calculate}}
$$\begin{array}{ll}
\displaystyle\int_0^1 \dfrac{1}{\sqrt{36+x^2}} \mathrm{d}x &= \displaystyle\int_0^1 \dfrac{1}{\sqrt{6^2+x^2}} \mathrm{d}x \\
&=\mathrm{arcsinh} \left( \dfrac{x}{6} \right) \Bigg|_0^1 \\
&= \mathrm{arcinh} \left( \dfrac{1}{6} \right) - \mathrm{arcsinh} (0) \\
&=\mathrm{arcsinh}\left( \dfrac{1}{6} \right).
\end{array}$$
\part[7] $\displaystyle\int_0^1 x^2+2x+1 \mathrm{d}x$ \\

\textit{Solution:} \textbf{This is of the type of integrals you could do in calc 1!} \\
{\color{blue} \href{http://www.wolframalpha.com/input/?i=integral+from+x%3D0+to+x%3D1+of+x%5E2%2B2x%2B1+dx}{Calculate}} 
$$\begin{array}{ll}
\displaystyle\int_0^1 x^2+2x+1 \mathrm{d}x &= \dfrac{x^3}{3} + x^2 + x \Bigg|_0^1 \\
&= \left( \dfrac{1}{3} + 1 + 1 \right) - 0 \\
&= \dfrac{7}{3}.
\end{array}$$
\part[7] $\displaystyle\int_0^1 \dfrac{1}{4-x^2} \mathrm{d}x$ \\
\textit{Solution:} \textbf{This problem resembles \texttt{\#}23 and \texttt{\#}86  in {\color{blue} \href{http://tomcuchta.com/teach/classes/2017/MATH-3315-Spring-2017-FairmontState/homework/hw3.php}{HW3}}}.\\
Note that $4=2^2$ and {\color{blue} \href{http://www.wolframalpha.com/input/?i=integral+from+x%3D0+to+x%3D1+of+1%2F(4-x%5E2)+dx}{calculate}}
$$\begin{array}{ll}
\displaystyle\int_0^1 \dfrac{1}{4-x^2} \mathrm{d}x &= \displaystyle\int_0^1 \dfrac{1}{2^2-x^2} \mathrm{d}x \\
&= \dfrac{1}{2} \mathrm{arctanh} \left( \dfrac{x}{2} \right) \Bigg|_0^1 \\
&=\dfrac{1}{2} \mathrm{arctanh} \left( \dfrac{1}{2} \right) - \dfrac{1}{2} \mathrm{arctanh}(0) \\
&=\dfrac{1}{2} \mathrm{arctanh} \left( \dfrac{1}{2} \right).
\end{array}$$
\part[7] $\displaystyle\int_1^2 \dfrac{1+\sin(\theta)}{\theta-\cos(\theta)} \mathrm{d}\theta$ \\
\textit{Solution:} \textbf{This problem resembles \texttt{\#}14, \texttt{\#}52, and \texttt{\#}55 in {\color{blue} \href{http://tomcuchta.com/teach/classes/2017/MATH-3315-Spring-2017-FairmontState/homework/hw1.php}{HW1}}}.\\
 Let $u=\theta-\cos(\theta)$ so that $\mathrm{d}u=1+\sin(\theta) \mathrm{d}\theta$. Moreover, if $\theta=1$ then $u=1-\cos(1)$ and if $\theta=2$ then $u=2-\cos(2)$. Now {\color{blue} \href{http://www.wolframalpha.com/input/?i=integral+from+theta%3D1+to+theta%3D2+of+(1%2Bsin(theta))%2F(theta-cos(theta))+dtheta}{calculate}}
$$\begin{array}{ll}
\displaystyle\int_1^2 \dfrac{1+\sin(\theta)}{\theta-\cos(\theta)} \mathrm{d}\theta &\stackrel{u=\theta-\cos(\theta)}{=} \displaystyle\int_{1-\cos(1)}^{2-\cos(2)} \dfrac{1}{u} \mathrm{d}u \\
&=\ln(|u|) \Bigg|_{1-\cos(1)}^{2-\cos(2)} \\
&=\ln(|2-\cos(2)|) - \ln(|1-\cos(1)|).
\end{array}$$
\part[7] $\displaystyle\int_e^{e^2} \dfrac{1}{x\ln(x)} \mathrm{d}x$ \\
\textit{Solution:} \textbf{This problem resembles \texttt{\#}14, \texttt{\#}52, and \texttt{\#}55 in {\color{blue} \href{http://tomcuchta.com/teach/classes/2017/MATH-3315-Spring-2017-FairmontState/homework/hw1.php}{HW1}}}.\\
Let $u=\ln(x)$ so that $\mathrm{d}u=\dfrac{1}{x} \mathrm{d}x$. Moreover, if $x=e$ then $u=\ln(e)=1$ and if $x=e^2$ then $u=\ln(e^2)=2$. Now {\color{blue} \href{http://www.wolframalpha.com/input/?i=integral+from+x%3De+to+x%3De%5E2+of+1%2F(x*ln(x))+dx}{calculate}}
$$\begin{array}{ll}
\displaystyle\int_e^{e^2} \dfrac{1}{x\ln(x)} \mathrm{d}x &\stackrel{u=\ln(x)}{=} \displaystyle\int_1^2 \dfrac{1}{u} \mathrm{d}u \\
&= \ln(|u|) \Bigg|_1^2 \\
&=\ln(2)-\ln(1) \\
&= \ln(2). 
\end{array}$$
\end{parts}

\end{questions}

\end{document}
