# Welcome to College Confidential!

## The leading college-bound community on the web

Join for FREE, and start talking with other members, weighing in on community discussions, and more.

Also, by registering and logging in you'll see fewer ads and pesky welcome messages (like this one!)

### As a CC member, you can:

• Post reviews of your campus visits.
• Find hundreds of pages of informative articles.
• Search from over 3 million scholarships.

# For anyone who's pretty good at calc.:

Registered User Posts: 1,991 Senior Member
edited January 2006
Integral(dx/((x-1)sqrt(4x^2-8x+3))
Post edited by KRabble88 on

## Replies to: For anyone who's pretty good at calc.:

• Registered User Posts: 1,183 Senior Member
Use (2x-2) = sec(theta)...
• Registered User Posts: 1,991 Senior Member
I'm sorry but I'm not following what you mean.
• Registered User Posts: 1,183 Senior Member
It's called trigonometric substitution. Usually if you have a quadratic expression inside a square root with no obvious derivative to that quadratic expression in your integrand, then you have to use trig substitution. I figured that if you had such an integral to solve you'd be familiar with the concept.
• Registered User Posts: 1,183 Senior Member
$\int_0^\infty S(x)dx$

And, unfortunately, LaTeX is not implemented on these forums. Oh well. No quick and easy help for you today.
• Registered User Posts: 1,991 Senior Member
Are you referring to the inverse trig function rules? This is as far as got in terms of trig. integration/derivation rules.

I tried completing the square, etc., but the form does not match so that I can put it into the arcsec form. Is there something I'm missing? Thanks!
• Registered User Posts: 1,991 Senior Member
Oh, thanks for your time though.

If you could, do you think you could name anything in particular that I may be missing? I don't need to see the actual problem drawn out.
• Registered User Posts: 803 Member
• Registered User Posts: 1,991 Senior Member
Forget it, I know what I did wrong. I added 4 to complete the square and I added ANOTHER 4 outside of the parenthesis instead of SUBTRACTING 4 outside of the parenthesis.
• Registered User Posts: 1,183 Senior Member
OK, the only arcsec you'll need to do is if you get "theta" (ok, let's just use t from now on) in the solution before resubstituting x into it.

So let's see. You have sqrt(4x^2 -8x +3) which you can write as sqrt(4x^2 -8x +4 -1) = sqrt( (2x-2)^2 -1).

Then you use the substitution (2x-2) = 1 * sec(t). You should draw a right triangle with (2x-2) as the length of the side adjacent to the angle "t" and 1 as the hypothenuse, for later use. The (x-1) part of your integral is (sec(t))/2 and also dx = (sec (t) tan (t) dt)/2.

So you have \int (sec t tan t dt)/(sec t sqrt (sec^2 t -1)) = \int (sec t tan t dt)/(sec t tan t) = \int dt = t+K (K some constant, unless I missed some definite integral bounds).

So you do have to arcsec your solution. t = arcsec(2x-2) (by arcsec of the original substitution). So the solution is arcsec (2x-2) +K (unless I made a mistake somewhere). No need for the triangle this time...
• Registered User Posts: 1,991 Senior Member
Alright, sorry if I wasted your time here. It turned out to be an incredibly stupid mistake (post #9). I was getting pretty frustrated and had no idea what I was doing wrong. But, thanks again, I truly, truly appreciate it.
This discussion has been closed.