`int_(5/2)^5(sqrt((25-x^2)^3))/(x^4)dx` is equal to (a) `pi/6` (b) `(2pi)/3` (c) `(5pi)/6` (d) `pi/3`

Options

A. `(pi)/6`
B. `(2pi)/3`
C. `(5pi)/6`
D. `(pi)/3`

Answer

D

Explanation

`I=int_(5//2) ^(5)(sqrt((25-x^(2))^(3)))/(x^(4)) dx`
Let `x=5 sin theta`
`:.dx=5 cos theta d theta`
`:. I=int_(pi//6)^(pi//2)(sqrt((25-25sin^(2) theta)^(3)))/(5^(4)sin^(4) theta)`
`=int_(pi//6)^(pi//2) (5^(3)cos ^(3) theta. 5 cos theta)/(5^(4)sin^(4) theta) d theta`
`=int_(pi//6)^(pi//2) cot^(2) theta (cosec^(2) theta-1)d theta`
`=int_(pi//6)^(pi//2) cot^(2) theta cosec^(2) theta d theta -int_(pi//6)^(pi//2) cot^(2) theta d theta`
`=int_(pi//6)^(pi//2) cot^(2) theta cosec^(2) theta d theta -int_(pi//6)^(pi//2) (cosec^(2) theta -1)d theta`
`=[-(cot^(3) theta)/3+cot theta + theta]_(pi//6)^(pi//2)`
`=-0+0+(pi)/2-(-(3sqrt(3))/3+sqrt(3)+(pi)/6)=(pi)/3`