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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.

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

3 Answers 3

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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?

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  • $\begingroup$ Thank you very much! $\endgroup$ Feb 9, 2020 at 21:03
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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.

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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

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