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

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

Vedclass · Accepted Answer

$\frac{8}{21}$

Vedclass · Answer

(C) Let $I = \int_0^{\pi / 2} \sqrt{\sin 	heta} \cos^3 	heta d 	heta$.Substitute $\sin 	heta = t$,then $\cos 	heta d 	heta = dt$.When $	heta = 0$,$t = 0$. When $	heta = \frac{\pi}{2}$,$t = 1$.We can rewrite $\cos^3 	heta d 	heta$ as $\cos^2 	heta (\cos 	heta d 	heta) = (1 - \sin^2 	heta) \cos 	heta d 	heta = (1 - t^2) dt$.Thus,$I = \int_0^1 \sqrt{t} (1 - t^2) dt = \int_0^1 (t^{1/2} - t^{5/2}) dt$.Integrating term by term: $I = \left[ \frac{t^{3/2}}{3/2} - \frac{t^{7/2}}{7/2} ight]_0^1$.$I = \left[ \frac{2}{3} t^{3/2} - \frac{2}{7} t^{7/2} ight]_0^1$.$I = \left( \frac{2}{3} - \frac{2}{7} ight) - (0 - 0) = \frac{14 - 6}{21} = \frac{8}{21}$.

$\int_0^{\pi / 2} \sqrt{\sin \theta} \cos ^3 \theta d \theta$ is equal to

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