# Quick ? Integrals

• May 10th 2007, 09:36 PM
qbkr21
Quick ? Integrals

Thanks!

-qbkr21
• May 10th 2007, 09:38 PM
Jhevon
Quote:

Originally Posted by qbkr21

Thanks!

-qbkr21

it's a trig identity:

sin(2x) = 2sinxcosx

Proof:
we apply the double angle formula for sine

sin2x = sin(x + x) = sinxcosx + sinxcosx = 2sinxcosx
• May 10th 2007, 10:39 PM
qbkr21
Re:
Re:
• May 10th 2007, 10:41 PM
Jhevon
Quote:

Originally Posted by qbkr21
Re:

the trick is to substitute 1 + cos^2x, so let u = 1 + cos^2x and things should work out nicely
• May 10th 2007, 10:43 PM
qbkr21
Re:
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, 10:45 PM
Jhevon
Quote:

Originally Posted by qbkr21
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?

-qbkr21

no, i meant at the begining. to get to the point you are now, it appears you let u = cos^2x. i'm saying go back and redo the problem from scratch using u = 1 + cos^2x

this is my 17:):)th post!!
• May 10th 2007, 10:51 PM
qbkr21
Re:
How did you tell it was 1+cos^2(x) from the beginning?
• May 10th 2007, 10:57 PM
Jhevon
Quote:

Originally Posted by qbkr21
How did you tell it was 1+cos^2(x) from the beginning?

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 11th 2007, 12:13 AM
CaptainBlack
Quote:

Originally Posted by qbkr21
Re:

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