# How do I Integrate This?

1. May 2, 2005

### Prototype

(x) sq. root of ((x^2) - 1)?

2. May 2, 2005

### Hessam

notice that the derivative of the inside is the same power as the one outside...

hmm, i'm thinking substitution... lol

let u = x^2 - 1
du = 2x dx
dx = du/2x

therefore you will then get (in terms of u)

1/2* (integral of sqrt(u))

$$1/2\int\sqrt{u}du$$

or somethign like that..

hope you can do the rest

PS, is this an ap calc question?

Last edited: May 2, 2005
3. May 2, 2005

### Zurtex

I would use the wonderful substiution:

$$x = \sec (u)$$

4. May 2, 2005

### Jameson

Or you could notate like this...

$$u = x^2-1$$

$$du = 2x dx$$

$$\frac{du}{2} = x dx$$

Last edited: May 3, 2005
5. May 2, 2005

### Hessam

can you explain why it is wrong?

i only know through calc bc, got the ap tommorow... if that explains it or not...

it would make sense to let dx = du/2x, so that i could just substitute it in

can't i treat dx and x as seperate variables?

6. May 2, 2005

### Data

I believe Jameson is just worried about your notation. du/(2x)=dx would be okay.

As well, Zurtex' method will work, but it's not necessary as the other suggestion in this thread is easier.

Last edited: May 2, 2005
7. May 2, 2005

### HallsofIvy

Staff Emeritus
It's not wrong, either way works:

xdx= du/2 so that $$x\sqrt{x^2-1)$$ becomes u1/2du/2 or

dx= du/2x so that $$x\sqrt{x^2-1}$$ becomes xu1/2du/2x and the x's cancel.

8. May 3, 2005

### dextercioby

I would suggest the delicious substitution

$$x=\cosh t$$

Daniel.

9. May 3, 2005

### Jameson

Ah, sorry, your notation does work. It just takes one extra step of cancelling the x's. I wasn't paying close enough attention. I fixed my post. Sorry for the messup.

Jameson