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

[Prototype]

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

 

[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))

[tex]1/2\int\sqrt{u}du[/tex]


or somethign like that..

hope you can do the rest

PS, is this an ap calc question?

 

[Zurtex]

I would use the wonderful substiution:

[tex]x = \sec (u)[/tex]

 

[Jameson]

Or you could notate like this...

[tex]u = x^2-1[/tex]

[tex]du = 2x dx[/tex]

[tex]\frac{du}{2} = x dx[/tex]

 

[Hessam]

That last step is a typo or it's incorrect.

It should be:

[tex]u = x^2-1[/tex]

[tex]du = 2x dx[/tex]

[tex]\frac{du}{2} = x dx[/tex]

Your end substitution is correct, with [tex]\frac{1}{2}\int\sqrt{u}du[/tex]
but that one step is off.

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 separate variables?

 

[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.

 

[HallsofIvy]

It's not wrong, either way works:

xdx= du/2 so that [tex]x\sqrt{x^2-1)[/tex] becomes u^1/2du/2 or

dx= du/2x so that [tex]x\sqrt{x^2-1}[/tex] becomes xu^1/2du/2x and the x's cancel.

 

[dextercioby]

I would suggest the delicious substitution

[tex]x=\cosh t [/tex]

Daniel.

 

[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