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int(0)^( pi)(sin x)/(1+cos^(2)x)dx=...

`int_(0)^( pi)(sin x)/(1+cos^(2)x)dx`=

Text Solution

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I = `int_0^pi (xsinx)/(1+cos^2x)dx`........(i)
also using `int_a^b f(x) dx = int_0^pi f(a+b-x) dx`
I = `int_0^pi( (pi-x)(sinx))/(1+cos^2x)dx`........(ii)
adding equation 1 and 2
2I = `pi int_0^pi (xsinx)/(1+cos^2x)dx` ...
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