For any `n in N`, the value of the intergral `int_(0)^(pi) (sin 2nx)/(sinx)dx` is,

Options

A. `pi`
B. `2pi`
C. `-pi`
D. none of these

Answer

D

Explanation

Let `I_(n)=underset(0)overset(pi)int(sin2nx)/(sinx) dx ` then,
`I_(n+1)=underset(0)overset(pi)int(sin2(n+1)x)/(sinx) dx`
`:.I_(n+1)-I_(n)=underset(0)overset(pi)int(sin(2n+2)x)/(sinx) dx.`
`rArr I_(n+1)-I_(n)=2underset(0)overset(pi)int(2sinx cos(2n+1)x)/(sinx) dx.`
`rArr I_(n+1)-I_(n)=2underset(0)overset(pi)intcos(2n+1)x dx`
`rArr I_(n+1)-I_(n)=(2)/(2n+1)[ sin (2n+1)x]_(0)^(pi)`
`rArr I_(n+1)-I_(n)=(2)/(2n+1)[ sin (2n+1)pi-sin o]=0`
`rArr I_(n+1)-I_(n)"for" n=1,2,3..`
`rArr I_(n+1)-I_(n-1)=...=I_(1)`
But, `I_(1)=underset(0)overset(pi)int(sin2x)/(sin x)dx= 2underset(0)overset(pi)int cos x dx=0`
Hence, `I_(n)=-0 "for all" n in N`.