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

Any help would be appreciated.

  • 1
    $\begingroup$ I take it you mean $\int \sin 2x \sqrt{(\sin^{2}x-cos^{3}x)}dx$. $\endgroup$
    – Autolatry
    Commented Mar 13, 2015 at 11:15
  • $\begingroup$ Yes I just edited it. $\endgroup$
    – user189013
    Commented Mar 13, 2015 at 11:16
  • $\begingroup$ The result is, according to Mathematica, both huge and non-elementary. $\endgroup$
    – mickep
    Commented Mar 13, 2015 at 12:00

1 Answer 1


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).

  • $\begingroup$ and after this substitution I have $\int 2u \sqrt {(1-u^2-u^3)}du$ what is the next step? $\endgroup$
    – user189013
    Commented Mar 13, 2015 at 11:37
  • 2
    $\begingroup$ 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. $\endgroup$ Commented Mar 13, 2015 at 11:39
  • $\begingroup$ Could you approximate this by integrating the Taylor expansion of the sqrt? $\endgroup$
    – ililil
    Commented Mar 13, 2015 at 11:52
  • 1
    $\begingroup$ This doesn't answer the question. $\endgroup$
    – abnry
    Commented Mar 13, 2015 at 12:11
  • $\begingroup$ I dont think this hint is helpful. You can't really reduce it any further..still $\endgroup$
    – MonK
    Commented Mar 13, 2015 at 13:15

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