TITLE: Odd $\sin/\cos$ integral
QUESTION [9 upvotes]: How to evaluate
$$\int \frac{\sin^3 x}{\cos^5x}dx\ ?$$
I've tried various substitutions with $\sin x = u$ or $\cos x = u$, I've tried using Euler's formula which result in too heavy calculations and I've tried using $\sin^2x + \cos^2x = 1$ in various forms without success.

REPLY [30 votes]: Another approach:
$$
\begin{align}
\int \frac{\sin^3 x}{\cos^5x}dx&=\int \frac{\sin^3 x}{\cos^3x\cos^2x}dx\\
&=\int\left(\frac{\sin x}{\cos x}\right)^3\frac{dx}{\cos^2x}\\
&=\int\tan^3x\ d(\tan x)\\
&=\frac14\tan^4x+C.
\end{align}
$$