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Question
Evaluate :`int_0^pi(xsinx)/(1+sinx)dx`
Solution
Let `I=int_0^pi(xsinx)/(1+sinx)dx`
`=int_0^pi((pi-x)sin(pi-x))/(1+sin(pi-x))dx [because int_0^a f(x)dx=int_0^af(a-x)dx]`
`=int_0^pi((pi-x)sinx)/(1+sinx)dx`
`=int_0^pi(pisinx)/(1+sinx)dx-I`
`I=int_0^pi(pisinx)/(1+sinx)dx-I`
`2I=int_0^pi(pisinx.(1-sinx))/((1+sinx)(1-sinx))dx`
`2I=int_0^pi(pisinx.(1-sinx))/(1-sin^2x)dx`
`(2I)/pi=int_0^pi(sinx.(1-sinx))/cos^2xdx`
`(2I)/pi=int_0^pi(sinx.-sin^2x)/cos^2xdx`
`(2I)/pi=int_0^pi(sinx)/cos^2xdx-int_0^pi(sin^2x)/cos^2xdx`
`(2I)/pi=int_0^pisecx.tanxdx-int_0^pitan^2xdx`
`(2I)/pi=[secx]_0^pi-int_0^pi(sec^2x-1)dx`
`(2I)/pi=[secpi-sec0]-int_0^pisec^2x.dx+int_0^pi1dx`
`(2I)/pi=[-1-1]-[tanx]_0^pi_[x]_0^pi`
`(2I)/pi=[-2]-[tanpi-tan0]+pi`
`(2I)/pi=[-2]-0+pi`
`thereforeI=((pi-2)pi)/2`
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