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Integral Calculus

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14 Nov 2009 14:56:30 IST

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

None

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

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aanchal nayyar

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14 Nov 2009 21:18:28 IST

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dx/(2+cosx)^2

putting cosx=2cos^(x/2)-1 in the given integral then the integral becomes

dx/(2cos^2(x/2)+1)^2 we can write 1 as sin^2(x/2)+cos^2(x/2) then the integral becomes

dx/(3cos^2(x/2)+sin^2(x/2))^2 and then further dividing numerator and denominator by cos^4(x/2) we obtain

sec^4(x/2)dx/(3+tan^2(x/2))^2 further putting tan^2(x/2) as t we have

(1+t^2 )dt/ (3+t^2)^2 now adding and subtracting 2 in numerator we obtain 2 integrals

dt/(3+t^2)-2dt/(3+t^2)^2 and now tese integrals can be easliy evaluated substitute the value of t in the final answer.

aanchal nayyar

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14 Nov 2009 21:19:42 IST

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and sorry could only think of method as yet but certainly come up with another method very soon.

taran

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14 Nov 2009 21:21:22 IST

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integration using differentiation

Dharav Solanki

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17 Nov 2009 18:23:24 IST

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Please read the post below

Dharav Solanki

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17 Nov 2009 18:24:51 IST

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Edited post.

Dharav Solanki

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17 Nov 2009 18:30:09 IST

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Sorry, I posted a wrong soltion. I will come back laters. Thank you!

Hari Shankar

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14 Mar 2010 11:30:43 IST

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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

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