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

Answer

`pi^(2)`

Explanation

Let `I=int _(-pi)^(pi)(2x(1+sinx))/(1+cos ^(2)x)dx`
`I=int_(-pi)^(pi)(2x)/(1+cos^(2)x)dx =int_(-pi)^(pi)(2x sinx)/(1+cos^(2)x)`
`rArr I=I_(1)_I_(2)` Now , `I _(1)= int_(-pi)^(pi) (2x)/(1+cos^(2)x)dx`
Let ` f(x) = (2x)/(1+cos ^(2)x)`
`rArr f(-x)= (-2x)/(1+cos ^(2)(-x))=(-2x)/(1+cos^(2)x)=-f(x)`
`rArr f(-x) =- f(x)` which shows that f (x) is an odd function.
`:. I_(1)=0`
Again . let ` g(x)= (2xsinx)/(1+cos^(2)x)`
`rArr g(-x)= (2(-x)sin(-x))/(1+cos^(2)(-x))= (2x sin x )/(1+ cos^(2)x)=g(x)`
`rArr g (-x) = g (x)` which shows that g (x) is an even function.
`:. I_(2)=int_(-pi)^(pi) (2x sin x)/(1+cos ^(2)x)dx = 2*2int _(0)^(pi)(x sinx)/(1+cos ^(2)x)dx`
`= 4 int _(0)^(pi) ((pi-x)sin(pi-x))/(1+[cos (pi-x)]^(2))dx = 4 int _(0)^(pi) ((pi-x)sinx)/(1+cos ^(2)x)dx`
`=4 int_(0)^(pi) (pisinx)/(1+cos^(2)x)dx -4 int _(0)^(pi)(xsinx)/(1+cos^(2)x)dx`
`rArr I_(2) = 4 pi int_(0)^(pi)(sinx)/ (cos ^(2)c)dx-I_(2)`
`rArr 2I_(2)=4piint _(0)^(pi) (sinx)/(1+cos^(2)x)dx`
Put `cos x= t rArr - sin x dx = dt`
`:. I_(2)=-2piint _(1)^(-1)(dt)/(1+t^(2))= 2 pi int _(-1)^(1)(dt)/(1+t^(2))=4pi int _(0)^(1)(dt)/(1+ t^(2))`
`=4pi [tan^(-1)t] _(0)^(1) =4pi [tan^(-1)1-tan^(-1)0]`
`4pi (pi//4-0)=pi^(2)`
`:. I-I_(1) =I_(2)=0+pi^(2)=pi^(2)`