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#1 2012-10-08 07:16:10
- 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 2012-10-08 07:26:16
- 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 2012-10-08 07:29:30
- zetafunc.
- Guest
Re: Integral
Never mind, forget W|A, it is unable to solve the problem so I do not trust its solution.
#4 2012-10-08 07:41:24
- bobbym
- Administrator
Online
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.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
#6 2012-10-08 08:06:24
- bobbym
- Administrator
Online
Re: Integral
Checking them now.
Yes, they are the same.
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
#7 2012-10-08 11:02:37
- anonimnystefy
- Real Member
Online
Re: Integral
Where can you get with the substitution x=(b-a)-a*cos(theta)?
Last edited by anonimnystefy (2012-10-08 11:02:52)
The limit operator is just an excuse for doing something you know you can't.
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