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Integral Calculus

Hot goIITian

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27 Jan 2009 19:53:14 IST

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integral from 0 to pie log(1-cos x) is it solved by formula or by properties?

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integral from 0 to pie log(1-cos x)is it solved by formula or by properties?

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10904him

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Joined: 29 Dec 2006

Posts: 191

27 Jan 2009 19:56:12 IST

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Properties are always the better option if available....Here

I = S ln(1-cosx)

I = S ln(1+cosx)

2I=S ln(sin^2x)

I = S lnsinx

Now solve this simple....

S---->Integral sign

dhirendrasinh

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Joined: 26 Jan 2009

Posts: 48

28 Jan 2009 00:22:25 IST

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Can we right log(int 0 to pi (1+cosx))

saharsha kumar keshkar

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Joined: 7 Aug 2008

Posts: 1104

28 Jan 2009 10:01:54 IST

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properties are better option.

Here u make less mistakes

Hari Shankar

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Joined: 28 Feb 2007

Posts: 2173

28 Jan 2009 15:27:00 IST

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$I = int_0^{pi} log (1 - cos x ) dx = int_0^{pi} log (1 +cos x ) dx \ \= rac{1}{2} int_0^{pi} log (sin^2 x ) dx \ \= int_0^{pi} log (sin x ) dx \ \= 2 int_0^{ rac{pi}{2}} log (sin x ) dx..............1 \ \= int_0^{ rac{pi}{2}} log (sin x ) dx + int_0^{ rac{pi}{2}} log (cos x ) dx \ \= int_0^{ rac{pi}{2}} log (sin 2x ) dx - int_0^{ rac{pi}{2}} log 2 dx \ \= rac{1}{2} int_0^{pi} log (sin t) dt - rac{pi}{2} log 2 \ \= int_0^{ rac{pi}{2}} log (sin t) dt - rac{pi}{2} log 2.........2$

From 1 and 2, we have

$2 int_0^{ rac{pi}{2}} log (sin t) dt = int_0^{ rac{pi}{2}} log (sin t) dt - rac{pi}{2} log 2 Rightarrow int_0^{ rac{pi}{2}} log (sin t) dt = - rac{pi}{2} log 2 \ \Rightarrow I = 2 int_0^{ rac{pi}{2}} log (sin t) dt = - pi log 2$

PS: @saharsha - apologies for not posting the solution earlier so that you could plagiarize it

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