1. Feb 27, 2010

### splelvis

from -2 to 2,
integral Ln(√(4-y^2)+2)dy,
how to integral that?

2. Feb 28, 2010

### Gib Z

This should be in the HOMEWORK forum. Have you even tried anything?

3. Feb 28, 2010

### Redbelly98

Staff Emeritus
Moderator's note: thread moved from "Calculus & Analysis"

4. Feb 28, 2010

### holezch

I'd like to know the answer to this as well. I started off by using parts:

$$yln (\sqrt{4 - y^{2}} + 2) + 1/2 {\int \frac{y^{2}}{4 - y^{2} + \sqrt{4-y^{2}}}}$$

and got this. For that integral I'm left with, I wanted to try decomposing it or use some kind of partial fractions method, but I have a sqrt expression in my denominator that has a variable inside.. I'm not sure what to do with that..
thanks

5. Feb 28, 2010

### snipez90

If you make the change of variables y = 2sin(x) in the original integral, this leaves you with an integrand of log(2cos(x) +2)cos(x). Then using integration by parts twice you should be able to get a solution.

holezch, your approach works as well I think. Try making the substitution I suggested. I haven't gone through with the calculations but I think it works.

6. Feb 28, 2010

### holezch

dang, that's a good idea.. thanks

7. Feb 28, 2010

### splelvis

is that should be log(2cos(x)+2)2cos(x)dx?
anyway that is good , thank you

8. Mar 1, 2010

### Gib Z

It's great how people just give out the answers to homework problems, to people who have not demonstrated even a minute of effort, especially when it is specifically in the forum guidelines to not do so.

And only one integration by parts is needed if you used a basic trigonometric identity, just for your information.

9. Mar 1, 2010

### snipez90

I'm guessing this was directed at me. First of all, I don't think giving the resulting integrand to an intermediate step of an integration problem which also involves a change of variables is a serious violation. Secondly, I had already read holezch's post before making my post, and the fact that he used integration by parts first made it difficult not to give the full approach. Still judging by the rest of the posts, I'm glad I was better able to help the one who at least tried something, but whatever.

And fine, only one integration by parts was probably needed. I didn't carry out any of the calculations on paper. If I see an approach that works, then I convey it.

10. Mar 1, 2010

### Gib Z

It doesn't matter if you think it didn't help him very much, Forum Guidelines state:

"NOTE: You MUST show that you have attempted to answer your question in order to receive help. You MUST make use of the homework template, which automatically appears when a new topic is created in the homework help forums. "

This is one of the reasons holezch shouldn't hijack someone elses homework help thread. Either wait for the answer to come through on the thread patiently, or give someone a personal message.

Physicsforums is becoming a haven for lazy students to get their homework done for them, enabled by occurrences like this, and Its becoming worse than ever. Is anyone else getting sick of this?