If `I_n=int_-pi^pi (sinnx)/((1+pi^x)sinx)dx, n=0,1,2, …,` then (A) `I_n=I_(n+2)` (B) `sum_(m=1)^10 I_(2m+1)=10pi` (C) `sum_(m=1)^10 I_(2m)=0` (D) `I_n=I_(n+1)`

Options

A. `I_(n)=I_(n+2)`
B. `sum_(m+1)^(10)I_(2m+1)=10pi`
C. `sum_(m=1)^(10)I_(2m)=0`
D. `I_(n)=I_(n+1)`

Answer

A, B, C

Explanation

Given ` I_(n) = int_(-pi)^(pi)(sinnx)/((1+pi^(x))sinx)dx` . . . (i)
Using `int_(a)^(b)f(x)dx = int_(a)^(b)f(b+a-x) dx` , we get
`I_(n) =int_(-pi)^(pi)(pi^(x)sinnx)/((1+pi^(x)) sin x)` .. . (ii)
On adding Eqs . (i) and (ii) , we have
`2I_(n)=int_(-pi)^(pi)(sin n x)/(sinx)dx =2 int_(0)^(pi)(sin nx)/(sinx)dx`
`[:' f(x) = (sin nx)/(sinx)` is an even function]
`rArr I_(n) = int_(0)^(pi)(sin nx)/(sinx)dx`
Now , `I_(n+2)-I_(n)= int _(0)^(pi) (sin (n+2)x- sin nx)/(sinx)dx`
`=int_(0)^(pi)(2cos(n+1)x*sinx)/(sinx)dx`
`=2int_(0)^(pi)cos (n+1) x dx =2 [(sin (n+1)x)/((n+1))]_(0)^(pi)=0` . . . (iii) `:. I_(n+2)=I_(n)`
Since , `I_(n) = int_(0)^(pi) (sinx)/(sin x)dx`
`rArr I_(1)=piandI_(2)=0`
From Eq . (iii) `I_(1)=I_(3)=I_(5)=....=pi`
and ` I_(2)=I_(4)=I_(6)=...=0`
`rArr sum_(m=1)^(10) I_(2m+1)=10piandsum_(m=1)^(10)I_(2m)=0`
`:.` Correct options are (a) , (b) , ( c) .