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\problemgrouptitle{Trigonometric Substitution}

\begin{problemblock}[2]{Evaluate the given integral.}

\item $\displaystyle \int \,\sqrt{16-x^2}\, dx$
\answer{
$\displaystyle 8 \sin^{-1}\! \left(\frac x4 \right)+\frac 12\, x\, \sqrt{16-x^2}+C$
}

\item $\displaystyle \int \,\frac{\sqrt{x^2+9}}x\, dx$
\answer{
$\displaystyle  \sqrt{9+x^2}+3\ln |\sqrt{9+x^2}-3 |-3\ln|x|+C$
}

\item $\displaystyle \int \,\frac1{x^4\, \sqrt{x^2-1}}\, dx$
\answer{
$\displaystyle  \frac{\sqrt{x^2-1}}x-\frac 13 \left( \frac{\sqrt{x^2-1}}x\right)^3+C$
}

\item $\displaystyle \int \,\frac 1 {\left(16-u^2\right)^{5/2}}\, dx$
\answer{
$\displaystyle  \frac 1{256}\left [\frac u{\sqrt{16-u^2}}+\frac 13 \left(\frac u{\sqrt{16-u^2}}\right)^3\right]+C$
}

\item $\displaystyle \int \,\frac1{\sqrt{4x^2-9}}\, dx$
\answer{
$\displaystyle  \frac 12 \ln \left  |2x+\sqrt{4x^2-9} \right |+C$
}

\item $\displaystyle \int \,\frac {x^2} {\left(x^2+4\right)^{3/2}}\, dx$
\answer{
$\displaystyle  \ln \left | \sqrt{4+x^2}+x \right |-\frac x{\sqrt{4+x^2}}+C$
}

\item $\displaystyle \int \,\frac {x^2} {\sqrt{5-3x^2}}\, dx$
\answer{
$\displaystyle  \frac 5{6 \sqrt 3}\, \sin^{-1}\! \left(\sqrt{\frac 35}\,x \right)-\frac 16\, x\, \sqrt{5-3x^2}+C$
}

\item $\displaystyle \int \,\frac {1} {x^2-4x+6}\, dx$
\answer{
$\displaystyle  \frac 1{ \sqrt 2}\, \tan^{-1}\! \left( \frac {x-2}{\sqrt 2} \right)+C$
}

\item $\displaystyle \int \,\frac x {\left(x^2-6x+15\right)^{5/2}}\, dx$
\answer{
$\displaystyle  -\frac 13 \left(x^2-6x+15 \right)^{-3/2}+\frac 1{12}\,\frac{x-3}{\sqrt{x^2-6x+15}}-\frac 1{36} \left(\frac{x-3}{\sqrt{x^2-6x+15}} \right)^3+C$
}

\item $\displaystyle \int \,\frac{3x+2} {\left(4-4x-x^2\right)^{3/2}}\, dx$
\answer{
$\displaystyle  \frac {4-x}{2\,\sqrt{4-4x-x^2}}+C$
}

\end{problemblock}

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