int_0^{pi /2} {sqrt {cos theta } {{sin }^3}th | vedclass

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(B) Let $I = \int_0^{\pi /2} \sqrt{\cos 	heta} \sin^3 	heta \, d	heta$. Substitute $t = \cos 	heta$,then $dt = -\sin 	heta \, d	heta$. When $	h

Vedclass · Accepted Answer

$\frac{8}{21}$

Vedclass · Answer

(B) Let $I = \int_0^{\pi /2} \sqrt{\cos 	heta} \sin^3 	heta \, d	heta$. Substitute $t = \cos 	heta$,then $dt = -\sin 	heta \, d	heta$. When $	heta = 0$,$t = 1$. When $	heta = \pi/2$,$t = 0$. Thus,$I = \int_1^0 \sqrt{t} (1 - t^2) (-dt) = \int_0^1 (t^{1/2} - t^{5/2}) \, dt$. Integrating term by term,we get $I = \left[ \frac{t^{3/2}}{3/2} - \frac{t^{7/2}}{7/2} ight]_0^1 = \left[ \frac{2}{3} t^{3/2} - \frac{2}{7} t^{7/2} ight]_0^1$. Evaluating at the limits,$I = \frac{2}{3} - \frac{2}{7} = \frac{14 - 6}{21} = \frac{8}{21}$.

$\int_0^{\pi /2} {\sqrt {\cos \theta } {{\sin }^3}\theta } \,d\theta = $

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