Integration with Trig Sub: Simplifying the Square Root

[jhahler]

Homework Statement

integral of sqrt(x^2-36)/x)

Homework Equations

sqrt(x^2-a^2) = asec(u)
Pythagorean identity

The Attempt at a Solution

I used trig sub on the x^2-36 and changed that to x=36sec(u) and dx= 36sec(u)tan(u). I simplified the square root in the numerator using sec^2(u)-1 = tan^2(u). This gave me (6tan(u)/36sec(u))(36sec(u)tan(u)) What trig identities do I need to proceed? Or did I use the wrong kind of trig sub? Any help is greatly appreciated in advance.

 

[tiny-tim]

hi jhahler!

This gave me (6tan(u)/36sec(u))(36sec(u)tan(u))

isn't that just tan² ?

 

[rock.freak667]

In your substitution, 'a' should be 6, not 36. Therefore your dx=6sec(u)tan(u).

After your change your numbers, and simplify, you will need to use the same trig identity you used in the beginning with tan^2(u).

 

[vanhees71]

I don't understand why you do such a complicated substitution. Perhaps I don't understand your notation right. It's way easier to use
$$x=6 \sin u.$$

 

[tiny-tim]

I used trig sub on the x^2-36 and changed that to x=36sec(u)

… $x=6 \sin u.$ …

sec² - 1 = tan²

sin² - 1 = minus cos²