For any `n in N, int_(0)^(pi) (sin (2n+1)x)/(sinx)dx` is equal to

Options

A. `pi`
B. 0
C. `n pi`
D. `(2n+1)=pi`

Answer

A

Explanation

We have,
`I_(n)= underset(0)overset(pi)int (sin (2n+1)x)/(sinx)dx`
`rArr I_(n+1)= underset(0)overset(pi)int (sin (2n+1)x)/(sinx)dx`
`:. I_(n+1)-I_(n)= underset(0)overset(pi)int (sin (2n+3)x-sin(2n+1)x)/(sinx)dx`
`rArr I_(n+1)-I_(n)= underset(0)overset(pi)int (2sin x cos (2n+2)x)/(sinx)dx`
`rArr I_(n+1)-I_(n)=(1)/(n+1)[ sin (2n+2)x]_(0)^(pi)=0`
`rArr I_(n+1)-I_(n)= I_(n-1)=...= I_(2)=I_(1)`
But, `I_(1)=underset(0)overset(pi)int(sin3x )/(sinx)dx=underset(0)overset(pi)int(3-4sin^(2)x)= 3pi-4xx(pi)/(2)=pi`
Hence, `underset(0)overset(pi)int(sin(2n+1)x)/(sinx) dx=pi`