Please seeRed

Thanks!

-qbkr21

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- May 10th 2007, 08:36 PMqbkr21Quick ? Integrals
Please see

**Red**

Thanks!

-qbkr21 - May 10th 2007, 08:38 PMJhevon
- May 10th 2007, 09:39 PMqbkr21Re:
Re:

- May 10th 2007, 09:41 PMJhevon
- May 10th 2007, 09:43 PMqbkr21Re:
Hmmm...for some odd reason I though you integrated first with "u" before you substitute "u" back into the problem, is this not always the case?

Ohh...I see I just substituted for cos(x) not 1+cos^2(x). I think that I made this a bit more difficult than it had to be.

-qbkr21 - May 10th 2007, 09:45 PMJhevon
- May 10th 2007, 09:51 PMqbkr21Re:
How did you tell it was 1+cos^2(x) from the beginning?

- May 10th 2007, 09:57 PMJhevon
the whole point of substitution is to turn a complicated integral into a simple one. the simpler we can make things the better. i realized that substituting u = cosx or u = cos^2x only would not really get rid of the problem, since i'd have to make a second substitution, say v = u^2 to get rid of the resulting u in the numerator, or would have to deal with a 1 + u in the denominator (not really a big deal, but still). now if i substituted 1 + cos^2x, i would get rid of the constant and the cos^2x, the derivative of a constant is zero, so it doesn't matter in the long run. and the integral will simply be -1/u, and you can't get any better than that

looking at the problem again, i saw that you just substituted cosx, but as you see, you can make life a lot simpler than that - May 10th 2007, 11:13 PMCaptainBlack
Unless you are required to use a particular method when confronted with

the integral:

I = 2* integral u/(1+u^2) du

the observation that 2u = d/du[1+u^2] should imeadiatly tell you that:

I = ln(1+u^2) + c.

The general idea is that:

integral [df/dx]/f(x) dx = ln(f(x) + c

RonL