Answer
$\frac{1}{2}\sec^2\phi+C$
Work Step by Step
$\int\frac{\sin\phi}{\cos^3\phi}\ d\phi$
Let $u=\cos \phi$. Then $du=-\sin\phi\ d\phi$, and $\sin\phi\ d\phi=-du$.
$=\int\frac{-1}{u^3}\ du$
$=-\int u^{-3}\ du$
$=-\frac{u^{-2}}{-2}+C$
$=\frac{1}{2u^2}+C$
$=\frac{1}{2\cos^2 \phi}+C$
$=\boxed{\frac{1}{2}\sec^2\phi+C}$
