plz integrate this tough one.........[1 over (2+cosx)whole square].....if possible in 2 or more met

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$int rac{1}{(2 + cos x)^2} dx = int rac{1}{(1 + 2 cos^2 rac{x}{2})^2} dx = int rac{sec^4 rac{x}{2} }{(2 + sec^2 rac{x}{2})^2} dx$

Now put t = an rac{x}{2}

Then the integral becomes $2 int rac{1+t^2}{(3+t^2)^2} dt$

This can be simplified as

$=2 int rac{1}{(3+t^2)} dt - 4 int rac{1}{(3+t^2)^2} dt$

First one is standard.

For the second one I will briefly outline the steps:

$int rac{1}{(3+t^2)^2} dt = int rac{1}{t^2( rac{3}{t}+t)^2} dt$

Now use

$rac{1}{t^2} = rac{1}{6} left[ dleft( rac{3}{t} +t ight) + dleft( rac{3}{t} -t ight) ight]$

and $left(3 + rac{1}{t} ight)^2 = left( rac{3}{t} -t ight)^2 + 12$

to finish off

Waiting for some quality posts in this site

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