The_ArtofScience said:
Homework Statement
I have a question dealing with this integral: \int\frac{x}{\sqrt{x^2-4}}. I did the trig substitution to check my other method of substitution and got two different answers: \sqrt{x^2-4} and \sqrt{x^2-4}/2. The latter is from trig substitution and the former from regular substitution.
The Attempt at a Solution
Ok, here is my work for the trig substitution
x=2sec@
dx=2sec@tan@
(x^2-4) = 4tan^2@
I finally get the integral of sec^2@ which gives me tan@ and looking from the triangle I drew its answer is \sqrt{x^2-4}/2. Why don't these two answers agree? Is there some rule suggesting not to use trig substitution or one way over the other?
Thank You
You skipped over the critical step! Yes, if you let x= 2 sec(\theta), then dx= 2sec(\theta)tan(\theta)d\theta[/itex] (you dropped the "d\theta" just as you dropped the "dx" in the original integral- bad habit.) and x^2- 4= 4tan^2(\theta) so \sqrt{x^2- 4}= 2 tan^2(\theta). Did you forget that the"4" in 4 tan^2(\theta) became "2" when you took the square root?
Now, the part you skipped- putting all that into the integral.
\int\frac{x dx}{\sqrt{x^2- 4}}= \int \frac{[2 sec(\theta)][2sec(\theta)tan(\theta)d\theta}{2 tan(\theta)}
= 2\int sec^2(\theta)d\theta= 2 tan(\theta)+ C= 2tan(sec^{-1}(\frac{x}{2})+ C
Notice the "2" still in there?
Now, if sec(\theta)= x/2, then we can represent that as a triangle with angle \theta, near side= 2, and hypotenuse= x. By the Pythagorean theorem, the opposite side has length \sqrt{x^2- 4} and so tan(\theta)= \sqrt{x^2- 4}/2. But because of the "2" multiplying the integral,
\int \frac{x dx}{\sqrt{x^2- 4}}= \sqrt{x^2- 4}+ C
exactly as you would get if you make the substitution u= x
2- 4.
