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\begin{document}

\begin{center}
\Huge{Trigonometric Substitution}
\end{center}


While working in a group make sure you:
\begin{packed_itemize}
\item Expect to make mistakes but be sure to reflect/learn from them!
\item Are civil and are aware of your impact on others.
\item Assume and engage with the strongest argument while assuming best intent.
\end{packed_itemize}


\begin{enumerate}
\item Let $x=3\sin(\theta)$.  Find the following in terms of $x$.

$\sin(\theta)$ \hspace{2in} $\theta$ \hspace{2in} $\cos (\theta)$

\vspace{2in}

\item Let $x=2\tan(\theta)$. Find the following in terms of $x$.

$\tan(\theta)$ \hspace{2in} $\cos(\theta)$ \hspace{2in} $\sec(\theta)$




\vfill
 %$\displaystyle{\int \cos^4(x)\sin(x) \,dx}$ \hspace{2.2in} 


\item Let $2x=3\tan(\theta)$. Find the following in terms of $x$.

$\tan(\theta)$ \hspace{2in} $\cos(\theta)$ \hspace{2in} $\theta$

\vfill



\newpage
\begin{center}
\begin{tabular}{ c c c c}
Expression & Substitution & Restrictions & Reason\\ 
\hline
&&&\\
$\sqrt{a^2-x^2}$ & $x=a\sin(\theta)$ & $\frac{-\pi}{2}\leq \theta \leq \frac{-\pi}{2}$ & $1-\sin^2(\theta)=\cos^2(\theta)$\\  
			& OR $x=a\cos(\theta)$ & $0\leq \theta \leq \pi$ & $1-\cos^2(\theta)=\sin^2(\theta)$\\ 
			\hline 
			&&&\\
$\sqrt{a^2+x^2}$ & $x=a\tan(\theta)$ & $\frac{-\pi}{2}\leq \theta \leq \frac{-\pi}{2}$ & $1+\tan^2(\theta)=\sec^2(\theta)$\\  
\hline
&&&\\
$\sqrt{x^2-a^2}$ & $x=a\sec(\theta)$ & $0\leq \theta \leq \pi$ & $\sec^2(\theta)-1=\tan^2(\theta)$\\  
\end{tabular}
\end{center}

\vspace{.5in}

\item Find  $\displaystyle{\int \frac{x}{(4x^2+9)^{\frac{3}{2}}\,dy}}$ 





\end{enumerate}

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