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

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

Question

(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 = $

Similar Questions

$\int\limits_0^3 \left( \frac{1}{\sqrt{x^2 + 4x + 4}} + \sqrt{x^2 - 4x + 4} \right) dx =$

The approximate value of $\int_1^3 \frac{dx}{2+3x}$ using Simpson's rule and dividing the interval $[1,3]$ into two equal parts is

Evaluate the definite integral $\int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x$.

$\int_{0}^{\pi} \sin x \, dx = $ . . . . . . .

For a real number $x$,let $[x]$ denote the greatest integer less than or equal to $x$. The smallest positive integer $n$ for which the integral $\int_{1}^{n} [x][\sqrt{x}] \, dx$ exceeds $60$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module