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?
Best Answer
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 $$ \int\frac{\cos(2x)}{\sin^{3/2}x}dx = 6E\left(\frac{\pi}{4}-\frac{x}{2}\; \bigg|\; 2\right) - \frac{2\cos x}{\sin^{1/2}x} $$