## Calculus: Early Transcendentals 8th Edition

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