`int_(0)^((pi)/(2))sqrt(costheta)*sin^(3)theta d theta=` . . .

Options

A. `(-20)/(21)`
B. `(-8)/(21)`
C. `(20)/(21)`
D. `(8)/(21)`

Answer

D

Explanation

Let `I=int_(0)^((pi)/(2))sqrt(costheta)*sin^(3)thetad theta`
`=int_(0)^((pi)/(2))sqrt(cos theta)*sin theta(1-cos^(2)theta)d theta`
Put `cos theta=t`
`implies-sintheta d theta=dt`
`implies sin theta d theta=-d t `
If `theta=0,t=1 and theta=(pi)/(2),t=0`
`thereforeI=int_(1)^(0)sqrt(t)(1-t^(2))(-d t)`
`=int_(0)^(1)(t^(1//2)-t^(5//2))dt`
`=(t^(3//2)/((3)/(2))-(t^(7//2))/((7)/(2)))_(0)^(1)=((2)/(3)t^(3//2)-(2)/(7)t^(7//2))_(0)^(1)`
`=((2)/(7)-(2)/(7))-(0-0)=(14-6)/(21)=(8)/(21)`