trigonometric integration problem

Question

Hello guys can someone please help me find the answer :

$$\int \sin x \tan x~dx$$

Answer 1

Note that $$\sin(x)\tan(x)=\frac{\sin^2(x)}{\cos(x)}=\frac{1-\cos^2(x)}{\cos(x)}$$

Answer 2

Bioche's rules say you should make the substitution $\;u=\sin x$, $\mathrm d u=\cos x\,\mathrm dx$, so that $$\int \sin x \tan x\,\mathrm dx=\int\frac{\sin^2x}{\cos x}\,\frac{\mathrm du}{\cos x}=\int\frac{u^2}{1-u^2}\,\mathrm du,$$ the integral of a rational function. Thus it comes down to a decomposition into partial fractions.