Any simple way for $\int \frac{\sqrt{\sin x-\sin^3x}}{\sqrt{1-\sin^3x}}dx$?

Question

I want to calculate the antiderivative of this function .

$\displaystyle f(x)=\frac{\sqrt{\sin x-\sin^3x}}{\sqrt{1-\sin^3x}}dx$

let $\sqrt {\sin x-sin^3x}=\sqrt{\sin x} .|\cos x|$ then I have :

$\displaystyle\int \frac{\sqrt{\sin x} .|\cos x|}{\sqrt{1-\sin^3x}}dx$ , then Assume $\cos x > 0$ and if i take $t= \sin x$ and then using inverse function of sin it will be complicated , then is there any simple method to Evaluate it ?

Answer 1

As you pointed out

$$\int\frac{\sqrt{\sin x-\sin^3x}}{\sqrt{1-\sin^3x}}dx$$

$$=\int\frac{\sqrt{\sin x(1-\sin^2x)}}{\sqrt{1-\sin^3x}}dx$$

$$=\int\frac{\sqrt{\sin x}\cos(x)}{\sqrt{1-\sin^3x}}dx$$

Let $u=\sin(x),$ then we get

$$=\int\frac{\sqrt{u}du}{\sqrt{1-u^3}}=\frac{2}{3}\sin^{-1}(u^{\frac{3}{2}})+C.$$