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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
Do your own homework.
• 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.