The value of int_{0}^{pi /2} (sqrt{sin theta} | vedclass

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(C) Let $I = \int_{0}^{\pi /2} (\sqrt{\sin 	heta} \cos 	heta)^3 d	heta = \int_{0}^{\pi /2} \sin^{3/2} 	heta \cos^3 	heta d	heta$.Using the Beta

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$8/45$

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(C) Let $I = \int_{0}^{\pi /2} (\sqrt{\sin 	heta} \cos 	heta)^3 d	heta = \int_{0}^{\pi /2} \sin^{3/2} 	heta \cos^3 	heta d	heta$.Using the Beta function formula $\int_{0}^{\pi /2} \sin^m 	heta \cos^n 	heta d	heta = \frac{\Gamma(\frac{m+1}{2}) \Gamma(\frac{n+1}{2})}{2 \Gamma(\frac{m+n+2}{2})}$,where $m = 3/2$ and $n = 3$:$I = \frac{\Gamma(\frac{3/2 + 1}{2}) \Gamma(\frac{3 + 1}{2})}{2 \Gamma(\frac{3/2 + 3 + 2}{2})} = \frac{\Gamma(5/4) \Gamma(2)}{2 \Gamma(13/4)}$.Since $\Gamma(2) = 1! = 1$ and $\Gamma(13/4) = \Gamma(1 + 9/4) = \frac{9}{4} \Gamma(9/4) = \frac{9}{4} \cdot \frac{5}{4} \Gamma(5/4)$:$I = \frac{\Gamma(5/4) \cdot 1}{2 \cdot (\frac{9}{4} \cdot \frac{5}{4} \Gamma(5/4))} = \frac{1}{2 \cdot \frac{45}{16}} = \frac{16}{90} = \frac{8}{45}$.

The value of $\int_{0}^{\pi /2} (\sqrt{\sin \theta} \cos \theta)^3 d\theta$ is

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