Let `n ge 2` be a natural number and `0 lt theta lt (pi)/(2)`, Then, `int ((sin^(n)theta - sin theta)^(1/n) cos theta)/(sin^(n+1) theta)d theta` is equal to (where C is a constant of integration)

Options

A. `(n)/(n^(2)-1)(1-(1)/(sin^(n+1)theta))^((n+1)/(n)+C`
B. `(n)/(n^(2)-1)(1+(1)/(sin^(n+1)theta))^((n+1)/(n)+C`
C. `(n)/(n^(2)-1)(1-(1)/(sin^(n+1)theta))^((n+1)/(n)+C`
D. `(n)/(n^(2)+1)(1-(1)/(sin^(n+1)theta))^((n+1)/(n)+C`

Answer

C

Explanation

Let `I = int((sin^(n)theta - sin theta)^(1//n) cos theta)/(sin^(n+1)theta)d theta`
Put `sin theta = t rArr cos theta d theta = dt`
`therefore I = int((t^(n)-t)^(1//n))/(t^(n+1))dt`
`=int([t^(n)(1-(t)/(t^(n)))]^(1//n))/(t^(n+1))dt`
`= int(t(1-1//t^(n-1))^(1//n))/(t^(n+1))dt = int((1-1//t^(n-1))^(1//n))/(t^(n))dt`
Put `1-(1)/(t^(n-1))=u`
or `1 - t^(-(n-1))=u rArr ((n-1))/(t^(n))dt = du`
`rArr" "(dt)/(t^(n))=(du)/(n-1)`
`rArr I = int(u^(1//n)du)/(n-1)=(u^(1/n+1))/((n-1)((1)/(n)+1))+C`
`=(n(1-(1)/(t^(n-1)))^((n+1)/(n)))/((n-1)(n+1))+C`
`(n(1-(1)/(sin^(n-1) theta))^((n+1)/(n)))/(n^(2)-1)+C" "[therefore u=1-(1)/(t^(n-1))and t = sin theta]`