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

Options

A. `pi/(2ab)`
B. `pi/(ab)`
C. `(pi^(2))/(2ab)`
D. `(pi^(2))/(ab)`

Answer

C

Explanation

Let `l=int_(0)^(pi)(xdx)/(a^(2)cos^(2)x+b^(2)sin^(2)x)" ...(i)"`
`rArr" "l=int_(0)^(pi)((pi-x)dx)/(a^(2)cos^(2)x+b^(2)sin^(2)x)" ...(ii)"`
On adding Eqs. (i) and (ii), we get
`2l=2piint_(0)^(pi//2)(dx)/(a^(2)cos^(2)x+b^(2)sin 2x)`
`rArr" "2l=2piint_(0)^(pi//2)(sec^(2)x)/(a^(2)+b^(2)tan^(@)x)`
Now, put `tanx=t rArr dx=(dt)/(sec^(2)x)`
`therefore" "2l=2piint_(0)^(oo)(dt)/(a^(2)+b^(2)t^(2))`
`=(1)/(b^(2))xx2piint_(0)^(oo)(dt)/((a^(2))/(b^(2))+t^(2))`
`=[(2pi)/(b^(2)).(b)/(a)tan^(-1).(bt)/(a)]_(0)^(oo)`
`=(2pi)/(ab)[tan^(-1)oo-tan^(-1)0]=(2pi)/(ab)xx(pi)/(2)`
`rArr" "l=(pi^(2))/(2ab)`