Answer
$$\frac{{{{\tan }^4}x}}{4} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{{\sin }^3}x}}{{{{\cos }^5}x}}dx} \cr
& {\text{rewrite the integrand}} \cr
& = \int {\frac{{{{\sin }^3}x}}{{{{\cos }^3}x}}\left( {\frac{1}{{{{\cos }^2}x}}} \right)dx} \cr
& {\text{using identities}} \cr
& = \int {{{\tan }^3}x{{\sec }^2}xdx} \cr
& {\text{substitute }}u = \tan x,{\text{ }}du = {\sec ^2}xdx \cr
& = \int {{{\tan }^3}x{{\sec }^2}xdx} \cr
& = \int {{u^3}} du \cr
& {\text{find the antiderivative}} \cr
& = \frac{{{u^4}}}{4} + C \cr
& {\text{ with}}\,\,\,u = \tan x \cr
& = \frac{{{{\tan }^4}x}}{4} + C \cr} $$