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How to compute the integral $\int_{0}^{\pi} (\sin \varphi)^{5 /3} d\varphi$? I tried to use $(\sin \varphi)^{5 /3} = (\sin \varphi)^{2} (\sin \varphi)^{-1/3}$. But it is not successful. Thank you very much.

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Note

\begin{align}\int_0^\pi (\sin \varphi)^{5/3}\, d\varphi &= \int_0^{\pi/2} (\sin \varphi)^{5/3}\, d\varphi + \int_{\pi/2}^\pi (\sin \varphi)^{5/3}\, d\varphi \\ &= \int_0^{\pi/2} (\sin \varphi)^{5/3}\, d\varphi + \int_0^{\pi/2} (\cos \varphi)^{5/3}\, d\varphi, \end{align}

where the transformation $\varphi \mapsto \pi/2 + \varphi$ was used in the last step. The last two integrals are Beta integrals of equal value $(1/2)B(4/3, 1/2)$. Now

$$B(4/3, 1/2) = \frac{\Gamma(4/3)\Gamma(1/2)}{\Gamma(4/3 + 1/2)} = \frac{\Gamma(4/3)}{\Gamma(11/6)}\sqrt{\pi}.$$

Therefore

$$\int_0^\pi (\sin \varphi)^{5/3}\, d\varphi = \frac{\Gamma(4/3)}{\Gamma(11/6)}\sqrt{\pi}.$$

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$$\int_{0}^{\pi}\left(\sin\varphi\right)^{5/3}\,d\varphi = 2\int_{0}^{\pi/2}\left(\sin u\right)^{5/3}\,du = 2\int_{0}^{1}\frac{v^{5/3}}{\sqrt{1-v^2}}\,dv $$ and: $$ 2\int_{0}^{1}\frac{v^{5/3}}{\sqrt{1-v^2}}\,dv = \int_{0}^{1} t^{1/3}(1-t)^{-1/2}\,dt =\frac{\Gamma(4/3)\Gamma(1/2)}{\Gamma(11/6)}$$ by exploiting the Euler Beta function. It follows that:

$$\int_{0}^{\pi}\left(\sin\varphi\right)^{5/3}\,d\varphi =\frac{\Gamma(1/3)\,\Gamma(1/6)}{5\sqrt{\pi}}.$$

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