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

Options

A. `(pi^(2))/(4)`
B. `pi^(2)`
C. zero
D. `(pi)/(2)`

Answer

B

Explanation

Let `l =int_(-x)^(pi) (2x)/(1+cos^(2)x)dx`
`+ 2int_(-pi)^(pi) (x sin x)/(1 + cos^(2) x)dx`
`rArr l = 0 + 4 int_(0)^(pi) ( x sin xdx)/(1 + cos^(2)x) " "…(i)`
`[because int_(-pi)^(pi) (2x)/(1+cos^(2)x)dx "is an odd function"]`
`rArr l = 4 int_(0)^(pi) ((pi-x)sin x)/(1 + cos^(2)x)dx" "...(ii)`
On adding Eqs. (i) and (ii), we get
`2l = 4pi int_(0)^(pi) (sin x)/(1 +cos^(2)x)dx`
Put `x = t rArr sin x dx = -dt`
`therefore l = 2pi int_(1)^(-1) (-dt)/(1+t^(2)) = 2pi [tan^(-1)t]_(-1)^(1)`
` = 2pi[(pi)/(4) + (pi)/(4)] = pi^(2)`