# 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: $\sin^2(x) = 1 - \cos^2(x)$.
– an4s
Feb 9, 2020 at 20:14
• There's an easier way to do this as outlined below, but the Weierstrass substitution always works. You only failed to apply it correctly. Feb 10, 2020 at 21:16

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?

• Thank you very much! Feb 9, 2020 at 21:03

Hint. Since $$\sin^2(x) = 1 - \cos^2(x)$$, the given integral can be written as $$\int{\frac{\sin(x)(1-\cos^2(x))}{2\cos^2(x)-3(1-\cos^2(x))}}dx$$ Now let $$t=\cos(x)$$ and it will turn out that new integrand function is a rational one which is easy to integrate.

integral(sin^3(x))/(2 cos^2(x) - 3 sin^2(x)) dx =

1/75 (15 cos(x) + 2 sqrt(15) tanh^(-1)(sqrt(5/3) - sqrt(2/3) tan(x/2)) + 2 sqrt(15) tanh^(-1)(sqrt(2/3) tan(x/2) + sqrt(5/3))) + constant