Chapter 4 - Integration - 4.5 Exercises - Page 301: 42

$\frac{1}{2}\sec^{2}x+C$

$\int \frac{sinx}{cos^{3}x}dx$ Let $u=cosx$ $u=cosx$ $\frac{du}{dx}=-sinx$ $dx=-\frac{1}{sinx}du$ $=-\int u^{-3}du$ $=-\frac{1}{2}u^{-2}+C$ $=\frac{1}{2}(\frac{1}{cos^{2}x})+C$ $=\frac{1}{2}sec^{2}x+C$