Problem source

Differentiate $\sec {t}$ with respect to $t$.
\begin{questionparts}
\item Use the substitution $x=\sec t$ to show that
$\displaystyle \int^2_{\sqrt 2} \frac{1}{ x^3\sqrt {x^2-1} } \; \mathrm{d}x
=\frac{\sqrt 3 - 2}{8} + \frac {\pi}{24} \;.$
\item Determine 
$\displaystyle \int \frac{1}  {( x+2)  \sqrt {(x+1)(x+3)} } \; 
\mathrm{d}x \;$.
\item Determine 
$\displaystyle \int \frac {1} {(x+2)  \sqrt {x^2+4x-5} } \; 
\mathrm{d}x \;$.
\end{questionparts}

Solution source

\[\frac{\d}{\d t} \left ( \sec t \right) = \frac{\sin t }{\cos^2 t} = \sec t \tan t \]

\begin{questionparts}
\item $\,$ \begin{align*}
&& I_1 &=  \int_{\sqrt{2}}^2 \frac{1}{x^3 \sqrt{x^2-1}} \\
x = \sec t, \d x = \sec t \tan t:&&&= \int_{t=\pi/4}^{t=\pi/3} \frac{1}{\sec^3 t \tan t} \sec t \tan t \d t \\
&&&=  \int_{t=\pi/4}^{t=\pi/3} \cos^2 t \d t \\
&&&=  \int_{t=\pi/4}^{t=\pi/3} \frac{1+\cos 2t}{2} \d t \\
&&&= \frac12 \frac{\pi}{12} + \frac12 \left (\sin \frac{\pi}{3} - \sin \frac{\pi}{4} \right) \\
&&&= \frac{\pi}{24} + \frac{\sqrt{3}-2}{8} \\
\end{align*}

\item $\,$ \begin{align*}
&& I_2 &= \int \frac{1}{(x+2)\sqrt{(x+1)(x+3)}} \d x \\
&&&= \int \frac{1}{(x+2)\sqrt{(x+2)^2-1}} \d x \\
&&&= \int \frac{1}{u\sqrt{u^2-1}} \d u \\
&&&= \sec^{-1} u + C \\
&&&= \sec^{1} (x+2) + C
\end{align*}

\item $\,$ \begin{align*}
&& I_3 &= \int \frac{1}{(x+2)\sqrt{(x+2)^2 - 9}} \d x \\
&&&= \int\frac{1}{9(\frac{x+2}{3})\sqrt{(\frac{x+2}3)^2 - 1}} \d x \\
u = \frac{x+2}{3}, 3\d u =\d x &&&= \frac19 \int \frac{1}{u\sqrt{u^2-1}} 3 \d u \\
&&&= \frac13 \sec^{-1} u + C \\
&&&= \frac13 \sec^{-1} \frac{x+2}{3} + C
\end{align*}
\end{questionparts}