So I need help integrating this: sqrt(1+cosx)dx

And integrating :cosxsinxdx on the interval 0 to 2pi.

Witht the second one I can only get to solutions that result in zero because the final always equals the initial.

Ideas?

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- Apr 5th 2011, 09:31 AMprofoundhelp proceeding further with a couple integrals
So I need help integrating this: sqrt(1+cosx)dx

And integrating :cosxsinxdx on the interval 0 to 2pi.

Witht the second one I can only get to solutions that result in zero because the final always equals the initial.

Ideas? - Apr 5th 2011, 09:37 AMprofound
I figured out the first one, I only need help with the second. Thanks

- Apr 5th 2011, 09:38 AMAckbeet
For the first problem, what ideas have you had so far?

For the second problem, have you considered that you might be getting the right answer? - Apr 5th 2011, 10:29 AMprofound
I used half angle identities and it worked with the first one. Is there an explanation as to why half angle identities work better sometimes?

For the second one I'm entirely sure its not zero. It goes from ds=3asinxcosxdx and the answer is 6a so somehow integrating sinxcosxdx should be 2.

Hmmm - Apr 5th 2011, 10:32 AMe^(i*pi)
Consider using a double angle identity

- Apr 5th 2011, 10:33 AMAckbeet
For the first problem, I would let $\displaystyle u=1+\cos(x),$ and go from there.

For the second problem, what do you get for the antiderivative? - Apr 5th 2011, 10:47 AMprofound
I have already with no luck

ds=3asinxcosxdx=(3a/2)sin(2x)dx from 0 to 2pi. Integrating that you ed up getting zero since the final is equal to the initial. - Apr 5th 2011, 10:51 AMAckbeet
So, what is that telling you?

- Apr 6th 2011, 05:38 AMprofound
It's not continuous? Not entirely sure what I should be doing from this point. It doesn't give me any help in my textbook that I can note. I have the answer its 6a however I cannot seem to avoid getting zero, no matter how I tackle it.

- Apr 6th 2011, 05:52 AMAckbeet
- Apr 6th 2011, 06:29 AMTheEmptySet
Just to add my 2 cents with definite integrals use a little geometery to see if you answer is reasonable.

Notice that $\displaystyle \displaystyle \sin(x)\cos(x)=\frac{1}{2}\sin(2x)$

This is a sine function with a peroid of Pi. Whenever you integrate since or cosine over one complete peroid you will get zero. Graph it.

So if you integrate over two peroids you will still get zero. - Apr 6th 2011, 08:10 AMprofound
The full question is "ds=3asinxcosxdx" where 3 and a are constants. I left then out because they have no affect on the answer. Anyways, I thought it was zero the whole time but the back of the book said 6a is the answer. Thanks!

- Apr 6th 2011, 08:12 AMAckbeet
Books have been known to be wrong in their answers before. I'd say the book's answer is wrong - that is, if you're sure that 6a is exactly what the book claims the answer is.

- Apr 6th 2011, 10:21 AMprofound
Haha, just thought I'd check. I strongly dislike when there is a wrong answer, especially when you are learning a new conept!

- Apr 6th 2011, 10:22 AMAckbeet