Answer:
I= integ.of dx/(4+5sin^2 x)
On dividing by cos^2x in Nr and Dr.
I=integ.of sec^2 x.dx/(4sec^2 x +5.tan^2 x)
I=integ.of sec^2 x.dx/(4+4tan^2 x+5tan^2 x)
I=integ.of sec^2 x.dx/(4+9tan^2 x)
Let tan x= t
sec^2 x.dx = dt
I=integ.of dt/(4+9t^2)
let t =(2/3)p , dt =(2/3)dp
I=integ.of (2/3).dp/(4+4p^2)
I=integ.of (2/12)dp/(1+p^2)
I=(1/6) tan^-1(p) +C
I=(1/6).tan^-1(3t/2) +C
I=(1/6).tan^-1{(3/2)tan x } +C , Answer.
