int_{0}^{frac{pi}{2}} sqrt{cos theta} cdot si | vedclass

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Question

(D) ધારો કે $I = \int_{0}^{\frac{\pi}{2}} \sqrt{\cos 	heta} \cdot \sin^{3} 	heta d 	heta$$= \int_{0}^{\frac{\pi}{2}} \sqrt{\cos 	heta} \cdot \sin

Vedclass · Accepted Answer

$\frac{8}{21}$

Vedclass · Answer

(D) ધારો કે $I = \int_{0}^{\frac{\pi}{2}} \sqrt{\cos 	heta} \cdot \sin^{3} 	heta d 	heta$$= \int_{0}^{\frac{\pi}{2}} \sqrt{\cos 	heta} \cdot \sin 	heta (1 - \cos^{2} 	heta) d 	heta$$\cos 	heta = t$ આદેશ લેતા,તેથી $-\sin 	heta d 	heta = dt$,અથવા $\sin 	heta d 	heta = -dt$.જ્યારે $	heta = 0, t = 1$ અને જ્યારે $	heta = \frac{\pi}{2}, t = 0$.$I = \int_{1}^{0} \sqrt{t} (1 - t^{2}) (-dt) = \int_{0}^{1} (t^{1/2} - t^{5/2}) dt$$= [\frac{t^{3/2}}{3/2} - \frac{t^{7/2}}{7/2}]_{0}^{1} = [\frac{2}{3} t^{3/2} - \frac{2}{7} t^{7/2}]_{0}^{1}$$= (\frac{2}{3} - \frac{2}{7}) - 0 = \frac{14 - 6}{21} = \frac{8}{21}$

$\int_{0}^{\frac{\pi}{2}} \sqrt{\cos \theta} \cdot \sin^{3} \theta d \theta = . . . . . .$

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