Need this answer quick please.

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.

1/5x+9 2/5y+7

step-by-step explanation:

answer: it is 1: 47 pm