Let `nge2` be a natural number and `0ltthetaltpi//2`. Then `int((sin^ntheta-sintheta)^(1//n)costheta)/(sin^(n+1)theta)d theta` is equal to (where C is a constant of integration)

Options

A. `(n)/(1- n^2) (1 - (1)/(cos^(n-1) theta) )^((n-1)/(n)) + C`
B. `(n)/(n^2 - 1) (1 - (1)/(cos^(n-1) theta) )^((n+1)/(n)) + C`
C. `(n)/(1 - n^2) ( 1 - (1)/(cos^(n-1) theta) )^((n-1)/(n)) + C`
D. None of these

Answer

A

Explanation

`I = int ((cos^n theta - cos theta)^(1/n) sin theta d theta)/( cos^(n+1) theta) `
` I = int ((cos^n theta - cos theta )/(cos^n theta) )^(1//n) . ( sin theta)/(cos^n theta) d theta`
`I = int ( 1 - (1)/(cos^(n-1) theta) )^(1//n) . (sin theta)/(cos^n theta) d theta`
let ` t = 1 - (1)/(cos^(n-1) theta)`
` dt = ( -(1-n))/(cos^n theta) xx - sin theta d theta , " " dt = (-(n - 1))/(cos^n theta) (sin theta) d theta`
`(dt)/(1-n) = (sin theta)/(cos^n theta) d theta therefore I = int t^(1//n) (dt)/(1 - n)`
`I = (1)/(1-n) (t^(1/n + 1))/(1/n + 1) + c , " " I = (n)/(1 - n^2) t^((n+1)//n) + c`
` I = (n)/(1 - n^2) (1 - (1)/(cos^(n-1) theta))^((n+1)//n) + c `