Evaluate the following :
`int_(0)^(pi)(x sinx)/(1+sinx)dx`

Explanation

Let `I=int_(0)^(pi)(x sinx)/(1+sinx)dx`
We use the property, `int_(0)^(a)f(x)dx=int_(0)^(a)f(a-x)dx.`
Here, `a=pi`. Hence changing x by `pi-x`, we get,
`I=int_(0)^(pi)((pi-x)sin(pi-x))/(1+sin(pi-x))dx`
`=int_(0)^(pi)((pi-x)sinx)/(1+sinx)dx=int_(0)^(pi)(pi sin x)/(1+sinx)dx-int_(0)^(pi)(x sinx)/(1+sinx)dx`
`therefore I=pi int_(0)^(pi)(sinx)/(1+sinx)dx-I`
`therefore 2I=pi int_(0)^(pi)(sinx)/(1+sinx).(1-sinx)/(1-sinx)dx`
`=pi int_(0)^(pi)((sinx(1-sinx))/(1-sin^(2)x))dx=pi int_(0)^(pi)(sinx-sin^(2)x)/(cos^(2)x)dx`
`=pi int_(0)^(pi)((sinx)/(cos^(2)x)-(sin^(2)x)/(cos^(2)x))dx=pi int_(0)^(pi)((1)/(cosx).(sinx)/(cosx)-tan^(2)x)dx`
`=pi int_(0)^(pi)(sec x tanx - sec^(2)x+1)dx=pi[sec x-tanx+x]_(0)^(pi)`
`=pi[(sec pi- tanpi+pi)-(sec 0- tan 0+0)]`
`=pi[(-1-0+pi)-(1-0+0)]pi(pi-1-1)=pi(pi-2)`
`therefore I=pi((pi-2)/(2))=pi((pi)/(2)-1)`