TITLE: Integrating $\int\frac{\sin^3x}{\cos x}\,\mathrm dx.$
QUESTION [9 upvotes]: How would I integrate the following:
.$$\int\frac{\sin^3x}{\cos x}\,\mathrm dx.$$
I am not sure what to do. I could split $\sin^3x=(1-\cos^2x)(\sin x)$
Then get $\int \tan(x)(1-\cos^2x)$
But would this be the right way to go.

REPLY [6 votes]: Actually you get $\int (\tan x-\sin x\cos x)dx$ which should be quite straight forward  
$$\int\frac{\sin^3x}{\cos x}dx = \int\frac{\sin x(1-\cos^2x)}{\cos x}dx = \int (\tan x - \sin x\cos x) dx = \int (\tan x - \frac{1}{2}\sin 2x)dx$$