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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $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 	he

Vedclass · Accepted Answer

$\frac{8}{21}$

Vedclass · Answer

(D) Let $I = \int_{0}^{\frac{\pi}{2}} \sqrt{\cos 	heta} \cdot \sin^{3} 	heta d 	heta$$= \int_{0}^{\frac{\pi}{2}} \sqrt{\cos 	heta} \cdot \sin 	heta (1 - \cos^{2} 	heta) d 	heta$Substitute $\cos 	heta = t$,then $-\sin 	heta d 	heta = dt$,or $\sin 	heta d 	heta = -dt$.When $	heta = 0, t = 1$ and when $	heta = \frac{\pi}{2}, t = 0$.$I = \int_{1}^{0} \sqrt{t} (1 - t^{2}) (-dt) = \int_{0}^{1} (t^{1/2} - t^{5/2}) dt$$= [\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}$

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

Similar Questions

Find $\int_{0}^{2}(x^{2}+1) dx$ as the limit of a sum.

$\int_{-1}^{1} |1 - x| \,dx = $

If $\int_{1}^{k}(3x^{2}+2x+1)dx=11$,then $k=$

$\int_0^\pi \frac{dx}{1 - 2a\cos x + a^2} = $

Let $[x]$ be the greatest integer function. Then, $\int_{-1}^{1} [x+2[x+2[x]]] dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module