Single Variable Calculus, Chapter 5, 5.5, Section 5.5, Problem 26

Find the indefinite integral $\displaystyle \int \sin t \sec ^2 (\cos t) dt$

If we let $\displaystyle u = \cos t$, then $\displaystyle du = - \sin t dt$, so $\sin t dt = - du$. And


$
\begin{equation}
\begin{aligned}

\int \sin t \sec^2 (\cos t) dt =& \int \sec^2 (\cos t) \sin t dt
\\
\\
\int \sin t \sec^2 (\cos t) dt =& \int \sec^2 u - du
\\
\\
\int \sin t \sec^2 (\cos t) dt =& - \int \sec^2 u du
\\
\\
\int \sin t \sec^2 (\cos t) dt =& - \tan u + C
\\
\\
\int \sin t \sec^2 (\cos t) dt =& - \tan (\cos t) + C

\end{aligned}
\end{equation}
$