Integrate √sin2x (sinx) - Limit 0 to pi/2

[Fernando Revilla]

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Help Integrate {√sin2x} (sinx) ? (underoot sin2x) * sinx.. Limit (0 to pi/2)?

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[Fernando Revilla]

We have: $$I:=\int_0^{\pi/2}\sqrt{\sin 2x}\sin x dx=\int_0^{\pi/2}\sqrt{2\sin x\cos x}\sin x dx=\sqrt{2}\int_0^{\pi/2}\sin^{3/2} x\cos^{1/2}x \; dx=\\
\sqrt{2}\frac{1}{2}B\left ( \frac{3/2+1}{2},\frac{1/2+1}{2} \right )=\frac{\sqrt{2}}{2}B\left ( \frac{5}{4},\frac{3}{4} \right )=\frac{\sqrt{2}}{2}\frac{\Gamma(5/4)\Gamma(1/4)}{\Gamma(2)}=\\
=\frac{\sqrt{2}}{2}\frac{(1/4)\Gamma((3/4)\Gamma(1/4)}{1!}=\frac{\sqrt{2}}{8}\Gamma(3/4)\Gamma (1/4)$$
Using the complement formula: $$I=\frac{\sqrt{2}}{8}\frac{\pi}{\sin (\pi/4)}=\frac{\pi}{4}$$