`int_(0)^((pi)/(2))sqrt(sin phi)cos^(5)phi d phi `

Explanation

माना `sin phi=t`
`rArr" "cos phi d phi = dt`
`phi = 0 "पर "t=sin 0=0`
`phi =(pi)/(2)" पर "t=sin.(pi)/(2)=1`
`therefore" "int_(0)^(pi//2)sqrt(sin phi).cos^(5)phi d phi`
`=int_(0)^(pi//2)sqrt(sin phi).(1-sin^(2)phi)^(2).cos phi d phi`
`=int_(0)^(1)sqrtt(1-t^(2))^(2)dt`
`=int_(0)^(1)(t^(1//2)-2t^(5//2)+t^(9//2))dt`
`=[(2)/(3)t^(3//2)-(4)/(7)t^(7//2)+(2)/(11)t^(11//2)]_(0)^(1)`
`=(2)/(3)-(4)/(7)+(2)/(11)`
`=(154-132+42)/(231)=(64)/(231)`