Solve Integration Problem: Exam in 2 Hrs - Help!

[UNIQNESS]

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

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.

 

[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 :/.

 

[UNIQNESS]

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

 

[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.

 

[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.

 

[UNIQNESS]

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

 

[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..

 

[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.

 

[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..

 

[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?

 

[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!