# Thread: help proceeding further with a couple integrals

1. ## help 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?

2. I figured out the first one, I only need help with the second. Thanks

3. 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?

4. 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

5. Consider using a double angle identity

6. 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?

7. 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.

8. So, what is that telling you?

9. 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.

10. Originally Posted by profound
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.
What is a? You don't mention a in your original problem statement anywhere. Are you sure you're comparing the right answer to the right problem? I would definitely claim that

$\displaystyle \displaystyle\int_{0}^{2\pi}\sin(x)\cos(x)\,dx=0.$

11. 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.

12. 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!

13. 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.

14. Haha, just thought I'd check. I strongly dislike when there is a wrong answer, especially when you are learning a new conept!

15. Originally Posted by profound
Haha, just thought I'd check. I strongly dislike when there is a wrong answer, especially when you are learning a new concept!
I can understand that, but you'd better get used to it: lots of books have lots of typos.