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#1 2012-10-07 08:16:10

zetafunc.
Guest

Integral

"(i) Use the substitution x = 2 - cosθ to evaluate the integral

.

(ii) Show that, for a < b,

where

and
."

I have done the first part and got

which is correct (according to WolframAlpha). But the second part of the question confuses me. I have done this:

Let x = (b - a) - cosθ, then dx = sinθdθ

and I have ended up with this:

but I do not know where to go from here. Help would be appreciated.

#2 2012-10-07 08:26:16

zetafunc.
Guest

Re: Integral

Hmm, I am skeptical about that last line. According to W|A it is giving me a horrendous-looking solution.

#3 2012-10-07 08:29:30

zetafunc.
Guest

Re: Integral

Never mind, forget W|A, it is unable to solve the problem so I do not trust its solution.

#4 2012-10-07 08:41:24

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 84,624

Re: Integral

Hi;

Alpha has a time limit. I do not think those two integrals are the same so something is wrong somewhere.


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

Online

#5 2012-10-07 08:50:36

zetafunc.
Guest

Re: Integral

These two integrals are identical...

#6 2012-10-07 09:06:24

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 84,624

Re: Integral

Checking them now.

Yes, they are the same.


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

Online

#7 2012-10-07 12:02:37

anonimnystefy
Real Member
From: The Foundation
Registered: 2011-05-23
Posts: 15,397

Re: Integral

Where can you get with the substitution x=(b-a)-a*cos(theta)?

Last edited by anonimnystefy (2012-10-07 12:02:52)


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