Question

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

Answer 1

`I=int_0^pix/(a^2cos^2x+b^2sin^2x)  dx.............(i)`

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

`I=int_0^pi(pi-x)/(a^2cos^2x+b^2sin^2x)dx...........(ii)`

`int_0^a f(x) dx = int_0^a f (a - x) dx`

Adding (i) and (ii), we get

`2"I" = int_0^pi (x + pi - x)/(a^2 cos^2 x + b^2 sin^2 x)  dx`

`2"I" = int _0^pi  pi/(a^2 cos^2 x + b^2 sin^2 x)  dx`

`2"I" = int_0^pi (pi sec^2 x )/(a^2 + b^2 tan^2 x)`     ........ `1/b^2 int_0^pi  (pi sec^2 x dx)/((a/b)^2 + tan^2 x)` 

`2"I" = pi/b^2 int  dt/(a/b)^2 + t^2`   .......... `[tan x = t  -> sec^2 x dx  = dt]`

`2"I" = pi/b^2 [(b/a) tan^-1 (bt/a)]_0^pi`

`2"I" = pi/(ab) [tan^-1 (b/a tan x)]_0^pi`

`2"I" = pi/(ab) (0 - 0) = 0`

2 I = 0

I = 0