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\title{\vspace*{-1.5in}\textbf{Math 2300-007: Trig. Substitution}\\\normalsize(Thanks to Faan Tone Liu)}
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%\underline{Math 1300-400: Quiz 1}
\lhead{Math 2300-007: Calculus II}
\chead{January 24, 2018}
\rhead{Trig. Substitution}
\vspace{-.5in}
\noindent
\textbf{Key Points:}
\begin{itemize} 
\item Use these substitutions when you see integrals with $\sqrt{a^2 - x^2}$ or $\sqrt{a^2+x^2}$.
\item Substitution for $\sqrt{a^2 + x^2}$:\vskip4cm
\item Substitution for $\sqrt{a^2 -x^2}$:\vskip4cm
\item It's worth remembering:\\[12pt]
$\displaystyle\int \frac{1}{a^2 + x^2}\,dx = \hspace{4.5cm} \int \frac{1}{\sqrt{a^2-x^2}}\,dx = $\\[12pt]
\item Other notes and tips:\vskip4cm
\end{itemize}
\newpage
Compute the following integrals:
\vskip12pt
\begin{enumerate}
 \item $\displaystyle \int \frac{1}{\sqrt{16+x^2}}\,dx$ \vskip9cm
 \item $\displaystyle  \int x^3\sqrt{1+x^2}\,dx$ \vskip9cm
\newpage
 \item $\displaystyle \int \frac{1}{y^2\sqrt{4-y^2}}\,dy$\vskip9cm
 \item $\displaystyle \int \frac{x^3}{\sqrt{9-x^2}}\,dx$ (What other method could you use?)\vskip9cm
\newpage
 \item $\displaystyle \int z\sqrt{1-z^2}\,dz$ (Hint: Can you do this another way?)\vskip6cm
 \item $\displaystyle \int \frac{1}{36+x^2}\,dx$ (Hint: Try some sneaky algebra first.)\vskip6cm
\item $\displaystyle \int \frac{1}{x^2 + 2x + 5}\,dx$ (Hint: Complete the square.)\vskip6cm
\end{enumerate}
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