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int0^pi(x dx)/(a^2cos^2x+b^2sin^2x)...

`int_0^pi(x dx)/(a^2cos^2x+b^2sin^2x)`

Text Solution

Verified by Experts

Given `I=int_0^pi (x)/(a^2dotcos^2x+b^2sin^2x)dx`...........(1)
Using the property `int_0^a(f(x))dx=int_0^af(a-x)dx`
`=int_0^pi (pi-x)/(a^2dotcos^2(pi-x)+b^2sin^2(pi-x))dx`
`=int_0^pi (pi-x)/(a^2dotcos^2x+b^2sin^2x)dx`.......(2)
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