Find the integral of (xsinx)/(1+(cosx)^2) from 0 to pi

use the identity:
integral of xf(sinx) dx from 0 to pi = pi/2 of the integral f(sinx) dx from 0 to pi
please explain your steps.

Here, f(sin x) = sin x/(1 + cos^2(x)), because we can rewrite sin x/(1 + cos^2(x)) as
sin x/(1 + (1 - sin^2(x)) = sin x/(2 - (sin x)^2)), which is in terms of sin x.
[So, f(t) = t/(2 - t^2).]
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So, ∫(x = 0 to π) x sin x dx/(1 + cos^2(x))
= ∫(x = 0 to π) x f(sin x) dx
= (π/2) ∫(x = 0 to π) f(sin x) dx, using the identity
= (π/2) ∫(x = 0 to π) sin x dx/(1 + cos^2(x))

Letting u = cos x, du = -sin x dx yields
(π/2) ∫(u = 1 to -1) -du/(1 + u^2)
= (π/2) ∫(u = -1 to 1) du/(1 + u^2)
= (π/2) arctan u {for u = -1 to 1}
= (π/2) ((π/4) - (-π/4))
= π²/4.

I hope this helps!