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

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Question

(C) माना $I = \int_0^{\pi / 2} \sqrt{\sin 	heta} \cos^3 	heta d 	heta$.प्रतिस्थापन: $\sin 	heta = t$,तब $\cos 	heta d 	heta = dt$.जब $	heta = 0

Vedclass · Accepted Answer

$\frac{8}{21}$

Vedclass · Answer

(C) माना $I = \int_0^{\pi / 2} \sqrt{\sin 	heta} \cos^3 	heta d 	heta$.प्रतिस्थापन: $\sin 	heta = t$,तब $\cos 	heta d 	heta = dt$.जब $	heta = 0$,तब $t = 0$. जब $	heta = \frac{\pi}{2}$,तब $t = 1$.यहाँ $\cos^3 	heta d 	heta = \cos^2 	heta (\cos 	heta d 	heta) = (1 - \sin^2 	heta) \cos 	heta d 	heta = (1 - t^2) dt$.अतः,$I = \int_0^1 \sqrt{t} (1 - t^2) dt = \int_0^1 (t^{1/2} - t^{5/2}) dt$.समाकलन करने पर: $I = \left[ \frac{t^{3/2}}{3/2} - \frac{t^{7/2}}{7/2} ight]_0^1$.$I = \left[ \frac{2}{3} t^{3/2} - \frac{2}{7} t^{7/2} ight]_0^1$.$I = \left( \frac{2}{3} - \frac{2}{7} ight) - 0 = \frac{14 - 6}{21} = \frac{8}{21}$.

$x$	$1$	$2$	$3$	$4$
$y$	$0.7111$	$0.7222$	$0.7333$	$0.7444$

$\int_0^{\pi / 2} \sqrt{\sin \theta} \cos ^3 \theta d \theta$ का मान ज्ञात कीजिए।

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