Answer
$$\frac{1}{3} \sec ^{3} x-\sec x+C $$
Work Step by Step
We integrate as follows:
\begin{align*}
\int \frac{\sin ^{3} x}{\cos ^{4} x} d x&=\int \frac{\sin ^{2} x \sin x}{\cos ^{4} x} d x\\
&=\int \frac{\left(1-\cos ^{2} x\right) \sin x}{\cos ^{4} x} d x\\
&=\int \frac{\sin x}{\cos ^{4} x} d x-\int \frac{\cos ^{2} x \sin x}{\cos ^{4} x} d x\\
&=\int \sec ^{3} x\tan x d x -\int \sec x \tan x d x\\
&=\int \sec ^{2} x \sec x \tan x d x-\int \sec x \tan x d x\\
&=\frac{1}{3} \sec ^{3} x-\sec x+C
\end{align*}
