If `int_(n)=int_(-pi)^(pi)(sin nx)/((1+pi^(x))sinx) dx, n=0,1,2,……….` then

Options

A. `I_(n)=I_(n+2)`
B. `underset(m=1)overset(10)sumI_(2m+1)=10pi`
C. `underset(m=1)overset(10)sumI_(2m)=0`
D. `I_(n)=I_(n+1)`

Answer

D

Explanation

Usingn `underset(-a) overset(a)int f(x) dx=underset(-a) overset(a)int{f(x)+f(x)}dx`, we have
`I_(n)=underset(-pi) overset(pi)int (sin n x)/((1+pi^(x))sinx)dx`
`rArr I_(n)=underset(0) overset(pi)int {(sin n x)/((1+pi^(x))sinx)+(sin nx)/((1+pi^(-x))sinx)}dx`
`rArr I_(n)=underset(0) overset(pi)int (sin n x)/(sin x) {(1)/(1+pi^(x))+(pi^(x))/(1+pi^(x))dx}`
`rArr I_(n)=underset(0) overset(pi)int(sin nx)/(sinx)dx`
`rArr I_(n+2)-I_(n)=underset(0) overset(pi)int(sin (n+2) x sin)/(sinx)dx`
`rArr I_(n+2)-I_(n)=2underset(0) overset(pi)intcos(n+1)x dx`
`rArr I_(n+2)-I_(n)=0 [ (sin (n+1)x)/(n+1)]_(0)^(pi)`
`rArr I_(n+2)-I_(n)=0`
`rArr I_(n+2) =I_(n)"for" n=0,1,2,3,...`
`rArr I_(n)=I_(n-2) =I_(n-4)=....= I_(2)=I(0)` if, n is even
and `L_(n)=I_(n-2)-I_(4)=....= I_(3),=I_(1)`, if n is odd
`rArr l_(n)={{:(0,"if n is even"),(pi, "n is odd"):}" "[ :. I_(0),0,I_(1)= underset(0)overset(pi)int (sin x)/(sin x) dx=pi]`
`rArr underset(m=1)overset(10)sum I_(2m)=0 and underset(m=1)overset(10)sum I_(2m+1)=10pi`