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\title{\vspace*{-1.5in}\textbf{Math 2300-007: Quiz 1}}
\date{}
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\begin{document}
\maketitle
%\underline{Math 1300-400: Quiz 1}
\lhead{Math 2300-013: Calculus II}
\chead{January, 19 2018}
\rhead{Quiz 1}
\vspace{-.5in}
\noindent
{Name: \underline{\phantom{Aalsdkjflaskdjflaskjflksdjflkdsjf}}} \hfill{Score: \underline{\phantom{lksdjflkdsjf}}}
\vskip1cm
\noindent
\begin{enumerate}
%\item (5 points) Compute the following derivatives
%\begin{enumerate}[(a)]
%\item $\displaystyle \frac{d}{dx} \int_3^x \cos(t)\,dt$\vskip2cm
%\item $\displaystyle \frac{d}{dx} \int_0^5 \ln(3t^2 + 1)\,dt$\vskip2cm
%\item $\displaystyle \frac{d}{dx} \int_{x^2}^{\sin{x}}\frac{1}{\sqrt{t^2 +t}}\,dt$\vskip2cm
%\end{enumerate}
\item (6 points) Evaluate the following\\[12pt]
\begin{tabular}{rl@{\hspace{1.5in}}rl}
 (a) & $\displaystyle\frac{d}{dx}\csc(x) = $ &(d)& $\displaystyle\int 3^x + x^2\,dx =$ \\[36pt]
 (b) & $\displaystyle\int\frac{1}{1 + t^2}\,dt = $ &(e) & $\ln(1) = $  \\[36pt]
 (c) & $\displaystyle\cos\left(\frac{7\pi}{6}\right)$= &(f) & $\displaystyle\frac{d}{d\theta}\sqrt{\sin(3\theta)}=$\\[36pt]
\end{tabular}
\item (1 point)  $u/du$-substitution is a technique for finding antiderivatives that is most closesly related to which rule for finding derivatives?\vskip.5in

\item (3 points) Evaluate the following integral. (Hint: Consider that $\sin^2(x) + \cos^2(x) =1$.)\vskip12pt
$
\displaystyle\int \cos^3(x)\sin^2(x)\,dx = 
$
%\begin{enumerate}[(a)]
%\item $\displaystyle \int \frac{\cos{\theta}}{1 + \sin^2{\theta}}\,d\theta$\vskip6cm
%\item $\displaystyle \int_1^4 \frac{\ln(x)\sqrt{\ln(x) + 1}}{x}\,dx$\vskip6.5cm
%\end{enumerate}

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\end{enumerate}

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