Answer
$\frac{\cos^2 x}{2}-\ln |\cos x|+C$
Work Step by Step
$\int\frac{\sin^3 x}{\cos x}dx$
$=\int\frac{\sin x\sin^2 x}{\cos x}dx$
$=\int\frac{\sin x(1-\cos^2 x)}{\cos x}dx$
Let $u=\cos x$. Then $du=-\sin x\ dx$, and $\sin x\ dx=-du$.
$=\int\frac{1-u^2}{u}*(-1)du$
$=-\int(\frac{1}{u}-\frac{u^2}{u})du$
$=-\int(\frac{1}{u}-u)du$
$=-(\ln |u|-\frac{u^2}{2}+C)$
$=-(\ln |\cos x|-\frac{\cos^2 x}{2}+C)$
$=\boxed{\frac{\cos^2 x}{2}-\ln |\cos x|+C}$