# Evaluating $\int \sin2x \sqrt{\sin^2x-\cos^3x} \, dx .$

I need some in evaluating the following integral: $$\int \sin2x \sqrt{\sin^2x-\cos^3x} \, dx .$$

Any help would be appreciated.

• I take it you mean $\int \sin 2x \sqrt{(\sin^{2}x-cos^{3}x)}dx$. Commented Mar 13, 2015 at 11:15
• Yes I just edited it. Commented Mar 13, 2015 at 11:16
• The result is, according to Mathematica, both huge and non-elementary. Commented Mar 13, 2015 at 12:00

Write $\sin 2x$ as $2 \sin x \cos x$, which suggests substituting $u = \sin x$ or $u = \cos x$. Before we do this, we must write the expression in the radical just in terms of $\sin$ or $\cos$: The latter is easier, thanks to the Pythagorean identity, $\sin^2 x = 1 - \cos^2 x$, and the resulting form suggests that we choose the substitution $u = \cos x$.
This reduces the integral to $$2 \int u \sqrt{1 - u^2 - u^3} du,$$ and using a CAS shows that the antiderivative involves the (nonelementary) elliptic integral functions (and are a serious mess otherwise too).
• and after this substitution I have $\int 2u \sqrt {(1-u^2-u^3)}du$ what is the next step? Commented Mar 13, 2015 at 11:37
• This is a good question ! I am just curious to see how it could be solved. By the way, are you sure of $\cos^3x$ ? With $\cos^2 x$ or $\cos^4 x$, I should be more comfortable. Commented Mar 13, 2015 at 11:39