Sign Up For Free

**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!)

- Reply to threads, and start your own.
- Post reviews of your campus visits.
- Find hundreds of pages of informative articles.
- Search from over 3 million scholarships.

KRabble88
Registered User Posts: **1,991** Senior Member

This discussion has been closed.

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

1,183Senior Member1,991Senior Member1,183Senior Member1,183Senior MemberAnd, unfortunately, LaTeX is not implemented on these forums. Oh well. No quick and easy help for you today.

1,991Senior MemberI 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!

1,991Senior MemberIf 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.

803Member1,991Senior Member1,183Senior MemberSo 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...

1,991Senior Member