Let I=∫π0xdxa2cos2x+b2sin2x ...(i)
I=∫π0π−xa2cos2(π−x)+b2sin2(π−x) ......[ By using ∫a0f(x)dx=∫a0f(a−x)dx]
I=∫π0π−xa2cos2x+b2sin2x⋯(ii)
Adding (i) and (ii), we get
2I=∫π0πa2cos2x+b2sin2xdx
⇒I=π2∫π0dxa2cos2x+b2sin2x
Dividing numerator and denominator by cos2x, we get
I=π2∫π0sec2xdxa2+b2tan2x
I=π∫π20sec2xdxa2+b2tan2x ......[ By using ∫2a0f(x)dx=2∫a0f(x)dx]
Putting btanx=t⇒bsec2 x dx=dt
The limits are, when x=0,t=0 and x=π2,t→∞
I=πb∫∞0dta2+t2=πb⋅1atan−1ta]∞0I=πab(tan−1∞−tan−10)=πab⋅π2⇒I=π22ab