65.9k views
11 votes
Need this answer quick please.​

Need this answer quick please.​-example-1
User Kpollock
by
8.5k points

1 Answer

7 votes

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.

User Gaurav Ram
by
7.9k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories