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I was trying to solve this indefinite integral; $$\displaystyle\int{\frac{\cos2x}{\sqrt{\sin^3 x}}}dx.$$

I tried using the online site; Integral Calculator. But it gave a very weird answer:

$$\int{\frac{\cos2x}{\sqrt{\sin^3 x}}}dx=2\operatorname{\Pi}\left(2\,;\dfrac{2x-{\pi}}{4}\,\middle|\,2\right)-4\operatorname{E}\left(\dfrac{2x-{\pi}}{4}\,\middle|\,2\right)+C$$ My attempt: \begin{align*} \int{\frac{\cos2x}{\sqrt{\sin^3 x}}}dx &=\int{\frac{1-2\sin^2x}{\sqrt{\sin^3 x}}}dx\\ &=\int{\frac{1-2\sin^2x}{\sin^{\frac{3}{2}}x}}dx\\ &=\int \left(\frac{1}{\sin^{\frac{3}{2}}x}-\frac{2\sin^2x}{\sin^{\frac{3}{2}}x}\right)dx\\ &=\int(\sin^{\frac{-3}{2}}x-2\sin^{\frac{1}{2}}x)dx\\ &=\int \sin^{\frac{-3}{2}}xdx-2\int \sin^{\frac{1}{2}}xdx. \end{align*}

What do I do now?

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    $\begingroup$ It's a bit strange, the mappings $x\mapsto \cos 2x$ and $x\mapsto \sin^{3}x$ can be negative simultaneously and that's a problem. $\endgroup$
    – A. P.
    Commented May 13, 2022 at 7:24
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    $\begingroup$ see math.stackexchange.com/questions/1469846/… $\endgroup$
    – sku
    Commented May 13, 2022 at 8:05
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    $\begingroup$ That $\Pi$ function can be simplified out to get $6E(\pi/4 + x/2 | 2) -\cos x/\sqrt{\sin x}$ but you're ending up with an elliptic integral regardless. The form of that antiderivative suggests a possible integration by parts. $\endgroup$ Commented May 13, 2022 at 12:33

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First, note that $d(\sin x\cos x)/dx = d(\sin(2x)/2)/dx = \cos(2x)$. So we can integrate by parts using $u = \cos^{3/2}x$ and $v = (\sin x \cos x)^{-1/2}$. This gives \begin{multline} \int\frac{\cos(2x)}{\sin^{3/2}x}dx = \int\cos^{3/2}x\frac{\cos(2x)dx}{(\sin x\cos x)^{3/2}}dx = -\frac{2\cos x}{\sin^{1/2}x}+\int\frac{2d\left[\cos^{3/2} x\right]}{\sin^{1/2}x\cos^{1/2}x}\\ = -\frac{2\cos x}{\sin^{1/2}x}+\int\frac{-3\sin x \cos^{1/2}x}{\sin^{1/2}x\cos^{1/2}x}dx = -\frac{2\cos x}{\sin^{1/2}x} - 3\int \sin^{1/2}xdx. \end{multline} This last integral can be transformed into an incomplete elliptic integral: $$ \int\sin^{1/2}xdx = \int \cos^{1/2}\left(\frac{\pi}{2}-x\right)d = \int\sqrt{1 - 2\sin^2\left(\frac{\pi}{4}-\frac{x}{2}\right)}dx = -2 E\left(\frac{\pi}{4}-\frac{x}{2}\; \bigg|\; 2\right). $$ So in total we have $$ \int\frac{\cos(2x)}{\sin^{3/2}x}dx = 6E\left(\frac{\pi}{4}-\frac{x}{2}\; \bigg|\; 2\right) - \frac{2\cos x}{\sin^{1/2}x} $$

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