Prerequisites :#(1): int_0^af(x)dx=int_0^af(a-x)dx#.
#(2): int_0^(2a)f(x)dx=2int_0^af(x)dx, if, f(2a-x)=f(x)#.
Let, #I=int_0^pix/(a^2sin^2x+b^2cos^2x)dx#.
Then, applying the above Rule (1), we get,
#I=int_0^pi(pi-x)/(a^2sin^2(pi-x)+b^2cos^2(pi-x))dx#,
#=int_0^pi(pi-x)/(a^2sin^2x+b^2(-cosx)^2)dx#,
#=int_0^pi(pi-x)/(a^2sin^2x+b^2cos^2x)dx#,
#=int_0^pipi/(a^2sin^2x+b^2cos^2x)dx-int_0^pix/(a^2sin^2x+b^2cos^2x)dx#.
#rArr I=int_0^pipi/(a^2sin^2x+b^2cos^2x)dx-I#,
#:. 2I=piint_0^pi1/(a^2sin^2x+b^2cos^2x)dx#,
To apply Rule (2), we verify, #f(2a-x)=f(x)#.
Here, #f(2a-x)=f(pi-x)=1/(a^2sin^2(pi-x)+b^2cos^2(pi-x))#,
#=1/(a^2sin^2x+b^2cos^2x)=f(x)#.
#:. 2I=2piint_0^(pi/2)1/{cos^2x(a^2tan^2x+b^2)dx#,
#:. I=piint_0^(pi/2)sec^2x/(a^2tan^2x+b^2)dx#.
We subst. #tanx=t rArr sec^2xdx=dt#.
Also, #x=0rArr t=tan0=0, &, x=pi/2, t to oo#.
#:. I=piint_0^oo1/(a^2t^2+b^2)dt,#
#=pi/a^2int_0^oo1/(t^2+(b/a)^2)dt#,
#=pi/a^2[1/(b/a)*arc tan (t/(b/a))]_0^oo#,
#=pi/(ab)[arc tan ((at)/b)]_0^oo#,
#=pi/(ab)[pi/2-0]#.
#rArr I=pi^2/(2ab)#.
