# Homework Help: Integration problem

1. Sep 1, 2005

### UNIQNESS

http://img.photobucket.com/albums/v295/uniqness/math.jpg [Broken]

I'm having a hard time trying to figure out how the last two steps were taken to obtain the answer. I actually have an exam in couple hours and I would really appreciate if someone could explain carefully as soon as possible. Thanks in advance.

Last edited by a moderator: May 2, 2017
2. Sep 1, 2005

### Nylex

In the first line, the x integration is being done first. The whole square root bit is a function of z and not x, so you just treat it like a constant. So, when you do the x integral, all you get is

$$x\sqrt{17 + 4z^2}$$

Next you have to put in the x limits: 0-1. Clearly, putting in x = 0 will give you nothing and putting in x = 1 gives you

$$\sqrt{17 + 4z^2}$$

This leaves you with the z integral at the end of the first line.

As for the second line, I'm a bit stuck right now :/.

Last edited: Sep 1, 2005
3. Sep 1, 2005

### UNIQNESS

lol thanks for the reply but the second line is what i'm having a problem with..

4. Sep 1, 2005

### arildno

The logarithmic expression is nothing else than the cosh or sinh inverse 8don't remember which)
In order to crack the integral, remember that $$Cosh^{2}y=1+Sinh^{2}y$$
Hence, make the change of varibles: $$z=a*Sinh(y)$$, where "a" is a constant to be fitted in the nicest manner possible.

5. Sep 1, 2005

### UNIQNESS

I still don't understand how the first line became the second line.. Also, how would i use the expression Cosh^2y = 1 + Sinh^2y there? I'd really appreciate if you could tell me step by step.. I just need to know how the first line became the second line.. This is not a homework problem as you can tell the answer is already there.. I'm just studying for a test.

6. Sep 1, 2005

### UNIQNESS

I know it's very early but is there anyone who can help me?

7. Sep 1, 2005

### arildno

OK, let's rewrite the integral in the first line as:
$$\int_{0}^{1}\sqrt{17+4z^{2}}dz=\sqrt{17}\int_{0}^{1}\sqrt{1+(\frac{2z}{\sqrt{17}})^{2}}dz$$
Thus, the change of variables $$\frac{2z}{\sqrt{17}}=Sinh(y)$$ is rather natural..

8. Sep 1, 2005

### UNIQNESS

Thank you for the help.. I guess I just don't remember the properties of sinh & cosh.. I thought there was an easier way to do it because I never used cosh or sinh to solve an integral.. I guess I will get this problem wrong if it comes out on the test.. Thanks though.

9. Sep 2, 2005

### UNIQNESS

my math exam is over so i don't need to know how to solve it any more but can anybody solve this thing step by step? when i usually post a problem, many people respond, but it seems like not many people know how to solve this..

10. Sep 2, 2005

### arildno

Well, you now have $$z=\frac{\sqrt{17}}{2}Sinh(y)$$, which implies that:
$$dz=\frac{\sqrt{17}}{2}Cosh(y)dy$$
Along with the new limits $$0\leq{y}\leq{Sinh}^{-1}(\frac{2}{\sqrt{17}})$$
Therefore, we get:
$$\sqrt{17}\int_{0}^{1}\sqrt{1+(\frac{2z}{17})^{2}}dz=\frac{17}{2}\int_{0}^{{Sinh}^{-1}(\frac{2}{\sqrt{17}})}Cosh^{2}ydy$$

All right?

11. Sep 2, 2005

### HallsofIvy

C'mon- in the very first post, Nylex told you how to go from the first line to the second- how the integration was done- and you replied "lol thanks for the reply but the second line is what i'm having a problem with..".
When it was pointed out that the second line was just a matter of evaluating functions you said "I still don't understand how the first line became the second line..", so we're back to what was said in the first post!