Evaluate `int_(0)^(1)(x^2/(1+x^2))dx`

Answer

B

Explanation

Let `I=int_(0)^(pi)(xsin^(2)x)/(1+2|cosx+sinx)dx` …………1
`=:.I=int_(0)^(pi)((pi-x)sin^(2)x)/(1+2|cosx|sinx) dx` …………….2
Adding 1 and 2 we get
`2I=pi int_(0)^(pi)(sin^(2)x)/(1+2|cos|sinx) dx`
`implies I=(pi)/2xx2int_(0)^((pi)/2)(sin^(2)x)/(1+2cosxsinx)dx`
`:.I=pi int_(0)^((pi)/2)(sin^(2)x)/(1+sin2x)dx`.............3
`:.I=piint_(0)^((pi)/2)(cos^(2)x)/(1+sin2x) dx`.................4
Adding 3 and 4 we get
`2I=pi int_(0)^((pi)/2)(dx)/((cosx+sinx)^(2))`
`=pi int_(0)^((pi)/2)(sec^(2)x)/((1+tanx)^(2))dx`
`=pi int_(0)^(oo) (dt)/(t^(2))dx` (putting `1+tanx=t`)
`=-pi(1/t)_(1)^(oo) =-pi (0-1)=pi`
Hence `I=pi//2`