For any `n in N, int_(0)^(pi) (sin^(2)nx)/(sin^(2)x)dx` is equal to

Options

A. `pi`
B. `npi`
C. 0
D. none of these

Answer

B

Explanation

`:. I_(n+1)-I_(n)= underset(0)overset(pi)int (sin^(2)(n+1)x-sin^(2)nx)/(sin^(2)x)dx` then,
` I_(n+1)= underset(0)overset(pi)int (sin^(2)(n+1)x)/(sin^(2)x)dx`
`:. I_(n+1)-I_(n)= underset(0)overset(pi)int (sin^(2)(n+1)x-sin^(2)nx)/(sin^(2)x)dx`
`rArr I_(n+1)-I_(n)= underset(0)overset(pi)int (sin^(2)(n+1)x)/(sin^(2)x)dx`
`rArr I_(n+1)-I_(n)=pi" " [ "See Example" 63]`
`:. I_(1),I_(2),..., I_(n),...` from an A.P. with common difference `pi`
`rArr I_(n) =pi+(n-1) pi=n pi" " [ :. I_(1)=underset(0)overset(pi)int(sin^(2)x)/(sin^(2)x)dx=pi]`