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Integrating Sec Tan^2

  • Thread starter TG3
  • Start date
  • #1
TG3
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0
Homework Statement
Integrate the square root of (x^2 -1) over the interval 0-1.

The attempt at a solution

First off, I know this is a quarter of a circle, but I'm not supposed to solve it that way.
Now then:

x = sec theta
dx= sec tan theta.

square root (x^2-1) dx
becomes
square root (sec^2-1) sec tan

Square root (tan^2) sec tan

tan sec tan

sec tan^2

This is as far as I can get- I can't see any u substitutions that would work here, and don't know this integral off the top of my head either.
 

Answers and Replies

  • #2
21
0
You went from (tan^2)sec tan to just (tan^2)sec, it should be (tan^3)sec. That doesn't matter though, think about the graph and what the area is doing on the interval (0,1).
 
  • #3
139
12
Hogger: There was a square root.

Integrals involving secants can be nasty; generally you have to integrate by parts. This is indeed the case with this integral, although first you should transform using identities to an integral only involving powers of secant. After integrating by parts once, you'll get an integral you've seen before, after which you can simply solve for it by algebraic means.
 
  • #4
33,169
4,854
Integrate the square root of (x^2 -1) over the interval 0-1.
First off, I know this is a quarter of a circle, but I'm not supposed to solve it that way.
The equation y = [itex]\sqrt{x^2 - 1}[/itex] is NOT the equation of a circle or any part of one.

The function above results in imaginary numbers for x in the interval (-1, 1). Are you sure that you have copied the problem correctly?
 

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