Question

and this prove that very urgent ​

Answer 1

Let

$I = \int\limits^\pi_0 {\frac{x}{a^2cos^2x+b^2sin^2x} } \, dx=>(i)$

$I = \int\limits^\pi_0 {\frac{(\pi - x)}{a^2cos^2(\pi-x)+b^2sin^2(\pi-x)} } \, dx$

$I = \int\limits^\pi_0 {\frac{(\pi-x)}{a^2cos^2x+b^2sin^2x} } \, dx=>(ii)$

Rest is in the attachment

HOPE THIS HELPS!!