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Question
Evaluate :
`∫_0^π(4x sin x)/(1+cos^2 x) dx`
Solution
`∫_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`
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