Evaluate: `int_(-pi)^pi(xsinx dx)/(e^x+1)`

Explanation

Let `I=pi_(-pi)^(pi)(x sin x dx)/(e^(x)+1)`……………..1
Replacing `x` by `0-x` or `r-x`, we get
`I=int_(-pi)^(pi)((-x)sin(-x)dx)/(e^(-x)+1)=int_(-pi)^(pi)(e^(x)xsin xdx)/(e^(x)+1)`…………….2
Adding 1 and 2 we get `2I=int_(-pi)^(pi) x sin x dx`
or `I=int_(0)^(pi)xsinxdx`
`int_(0)^(pi)(pi-x)sin(pi-x)dx=int_(0)^(pi)x sin x dx-I` or `I=pi`