Solution
= Solution
Put $u=\cos x$, $du=-\sin x\,dx$, and use $\sin^2x=1-u^2$. Then
$$
\int\frac{\sin^3x}{\cos^4x}dx=-\int\frac{1-u^2}{u^4}du=\frac1{3u^3}-\frac1u+C.
$$
Therefore
$$
\boxed{\int\frac{\sin^3x}{\cos^4x}dx=\frac1{3\cos^3x}-\frac1{\cos x}+C.}
$$
This holds on intervals where $\cos x\ne0$.