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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
From: Bumpkinland
Registered: 2009-04-12
Posts: 107,123

### 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.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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

zetafunc.
Guest

### Re: Integral

These two integrals are identical...

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

bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 107,123

### Re: Integral

Checking them now.

Yes, they are the same.

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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

anonimnystefy
Real Member
From: Harlan's World
Registered: 2011-05-23
Posts: 16,018

### 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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