The value of `int_(-pi)^(pi)(2x(1+sinx))/(1+cos^(2)x)dx` is

Options

A. `pi`
B. `pi^(2)`
C. `2pi^(2)`
D. `pi^(2)//2`

Answer

B

Explanation

`I=int_(-pi)^(pi)(2x(1+sinx))/(1+cos^(2)x)dx`
`=int_(-pi)^(pi)(2x)/(ubrace(1+cos^(2)x)_("odd function"))dx+2int_(-pi)^(pi)(x sin x)/(1+cos^(2)x)dx`
`=0+4int_(0)^(pi)(x sinx)/(1+cos^(2)x)dx`
`=4int_(0)^(pi) ((pi-x)sinx)/(1+cos^(2)x)dx`
`=4pi int_(0)^(pi)(sinx)/(1+cos^(2)x)dx -4int_(0)^(pi)(x sinx)/(1+cos^(2)x)dx`
`implies 2I=4pi int_(0)^(pi)(sinx)/(1+cos^(2)x)dx`
`implies I=2pi int_(0)^(pi) (sinx)/(1+cos^(2)x)dx`
`=-2pi [tan^(-1)(cosx)]_(0)^(pi)`
`=-2pi xx(-(pi)/2)=pi^(2)`