Question

Evaluate :

`∫_0^π(4x sin x)/(1+cos^2 x) dx`

Answer 1

`∫_0^π(4x sin x)/(1+cos^2 x) dx..................(1)`

Using f (x) = f (a−x), we get:

`I=∫_0^π(4(pi-x) sin x)/(1+cos^2 x) dx .....................(2)`

Adding (1) and (2), we get:

`2I=4int_0^pi(pi sinx)/(1+cos^2x)dx`

`I=2int_0^pi(pi sinx)/(1+cos^2x)dx`

Let cos x=t.

⇒−sin xdx=dt

`⇒I=2π∫_1^(−1)−1/(1+t^2)dt`

`=>I=-2pi tan^(-1) t_1^(-1)`

`=>I=-2pi(-pi/4-pi/4)`

`=>I=pi^2`