`intsin^(3)x.cos^(2)xdx` is equal to

Options

A. `(sin^(5)x)/(5)-(sin^(3)x)/(3)+C`
B. `(sin^(5)x)/(5)+(sin^(3)x)/(3)+C`
C. `(cos^(5)x)/(5)-(cso^(3)x)/(3)+C`
D. `(cos^(5)x)/(5)+(cos^(3)x)/(3)+C`

Answer

C

Explanation

Let `l=int(1-cos^(2)x) cos^(2)x. sinx x`
Put `cos x = t rArr - sin x dx=dt`
`therefore" "l=- int(1-t^(2))t^(2)dt=int(t^(4)-t^(2))dt`
`=(t^(5))/(5)-(t^(3))/(3)+C`
`=((cosx)^(5))/(5)-((cosx)^(3))/(3)+C`