Integral $\int{\frac{\sin^3(x)}{2\cos^2(x)-3\sin^2(x)}}dx$ I want to integrate the following expression:
$$\int{\frac{\sin^3(x)}{2\cos^2(x)-3\sin^2(x)}}dx$$
I tried using a $t=\tan(\frac{x}{2})$ substitution but the terms were not cancelled. Then I tried using $1=\cos^2(x)+\sin^2(x)$ but couldn't get anywhere. I hope some of you can help, thanks. 

Hint:
Recall that $$\sin^2(x) = 1 - \cos^2(x)$$
Therefore,
$$\int\dfrac{\sin^3(x)}{2\cos^2(x) - 3\sin^2(x)}\,\mathrm dx\equiv\int\dfrac{1 - \cos^2(x)}{5\cos^2(x) - 3}\sin(x)\,\mathrm dx$$
Let $u = \cos(x)\implies\mathrm du = -\sin(x)\mathrm dx$. So,
$$\int\dfrac{1 - \cos^2(x)}{5\cos^2(x) - 3}\sin(x)\,\mathrm dx\equiv\int\dfrac{u^2 - 1}{5u^2 - 3}\,\mathrm du = \dfrac15\int\mathrm du - \dfrac25\int\dfrac1{5u^2 - 3}\,\mathrm du$$
Can you take it from here?
