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Evaluate: int0^pi(xsinx)/(1+cos^2x)\ dx...

Evaluate: `int_0^pi(xsinx)/(1+cos^2x)\ dx`

Text Solution

Verified by Experts

The correct Answer is:
`pi^(2)//4`

Let `I=int_(0)^(pi)(xsinxdx)/(1+cos^(2)x)`…………..1
or `I=int_(0)^(pi)((pi-x)sinx dx)/(1+cos^(2)x)`…………2
Adding 1 and 2 we get
`2I=piint_(0)^(pi)(sinxdx)/(1+cos^(2)x)`
or `I=-(pi)/2 int_(1)^(-1)(dt)/(1+t^(2))=-(pi)/2[tan^(-1)t]_(1)^(-1)`
[Putting `cosx=t, -sinx dx=dt`]
`=-1/2[tan^(-1)(-1)-tan^(-1)]=pi^(2)//4`
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