Trig substitution problem

[ShatteredMirror]

Hi, I've been working on a trig substitution problem and I cant for the life of me figure out how to finish it. I'll post the progress I've made so far on the problem.

\(\displaystyle \large \int\frac{\sqrt{x^2-36}}{x^2}dx\)

\(\displaystyle \large x = 6 \sec \theta \)
\(\displaystyle \large dx = 6 \sec \theta \tan \theta d\theta\)

\(\displaystyle \large \sqrt{x^2-36} = 6 \tan \theta \)

\(\displaystyle \large \int\frac{6 \tan \theta \cdot 6 \sec \theta \tan \theta}{(6 \sec \theta)^2} d\theta = \int\frac{36 \sec \theta \tan^2\theta}{36 \sec^2\theta} d\theta\)
36's cancel, \(\displaystyle \sec \theta\) cancels.
\(\displaystyle \large \int\frac{\tan^2\theta}{\sec \theta} = \int\frac{\frac{\sin^2\theta}{\cos^2\theta}}{\frac{1}{\cos \theta}} \cdot \frac{\cos \theta}{\cos \theta}\)
\(\displaystyle \cos \theta\) cancels from top and bottom and we're left with...
\(\displaystyle \large \int\frac{\sin^2\theta}{\cos \theta}\)

But where should I go from here?
I thought i *might* be able to change it to..
\(\displaystyle \large \int\frac{\sin \theta}{\cos \theta} \cdot \sin \theta = \int \tan \theta \cdot \sin \theta \)

but is that correct?
Where should I go from here?!
Thank you for your time!

EDIT: Nevermind guys, I figured it out!

from the step:
\(\displaystyle \large \int\frac{\tan^2 \theta}{\sec \theta} d\theta \)
you use the trig identity for \(\displaystyle \tan^2x\)
\(\displaystyle \large \int\frac{\sec^2 \theta - 1}{\sec \theta} = \int\frac{\sec^2\theta}{\sec\theta} d\theta - \int\frac{1}{\sec \theta} d\theta \)
cancel the \(\displaystyle \sin \theta \) from the top and bottom.
\(\displaystyle \large \int \sec \theta d\theta - \int \cos \theta d \theta \)
Finally integrate to get...
\(\displaystyle \large \ln{| \sec \theta + \tan \theta|} - \sin \theta + C \)
\(\displaystyle \large \ln{|\frac{x}{6} + \frac{\sqrt{x^2 - 36}}{6}|} - \frac{\sqrt{x^2-36}}{x} + C \)

 

[lookagain]

\(\displaystyle \large \int\frac{\sqrt{x^2-36}}{x^2}dx\)

\(\displaystyle \large x = 6 \sec \theta \)

\(\displaystyle \large dx = 6 \sec \theta \tan \theta d\theta\)

\(\displaystyle \large \sqrt{x^2-36} = 6 \tan \theta \)

\(\displaystyle \large \int\frac{6 \tan \theta \cdot 6 \sec \theta \tan \theta}{(6 \sec \theta)^2} d\theta = \int\frac{36 \sec \theta \tan^2\theta}{36 \sec^2\theta} d\theta\)

36's cancel,

\(\displaystyle \sec \theta\) cancels.

\(\displaystyle \large \int\frac{\tan^2\theta d\theta}{\sec \theta } = \int\frac{\frac{\sin^2\theta}{\cos^2\theta}}{\frac{1}{\cos \theta}} \cdot \frac{\cos \theta d\theta}{\cos \theta}\)

\(\displaystyle \cos \theta\)

cancels from top and bottom and we're left with...

\(\displaystyle \large \int\frac{\sin^2\theta d\theta}{\cos \theta}\)

But where should I go from here?I thought i *might* be able to change it to...

\(\displaystyle \large \int\frac{\sin \theta}{\cos \theta} \cdot \sin \theta d\theta = \int \tan \theta \cdot \sin \theta d\theta \)

but is that correct?Where should I go from here?!Thank you for your time!

EDIT: Nevermind guys, I figured it out!from the step:

\(\displaystyle \large \int\frac{\tan^2 \theta}{\sec \theta} d\theta \)

you use the trig identity for \(\displaystyle \tan^2x\)

\(\displaystyle \large \int\frac{\sec^2 \theta - 1}{\sec \theta}d\theta = \int\frac{\sec^2\theta}{\sec\theta} d\theta - \int\frac{1}{\sec \theta} d\theta \)

cancel the \(\displaystyle \sec \theta \)

from the top and bottom.



\(\displaystyle \large \int \sec \theta d\theta - \int \cos \theta d \theta \)

Finally integrate to get...

\(\displaystyle \large \ln{| \sec \theta + \tan \theta|}\ - \ \sin \theta \ + \ C \)



\(\displaystyle \large \ln{\bigg|\frac{x}{6} + \frac{\sqrt{x^2 - 36 \ }}{6}\bigg|} \ - \ \frac{\sqrt{x^2 - 36 \ }}{x} \ + \ C \)

\(\displaystyle \large \ln{\bigg|\frac{x \ + \ \sqrt{x^2 - 36 \ }}{6} \bigg|} \ - \ \frac{\sqrt{x^2 - 36 \ }}{x} \ + \ C_1 \)

\(\displaystyle \ln\bigg|x \ + \ \sqrt{x^2 - 36 \ }\bigg| \ - \ \ln(6) \ - \ \dfrac{\sqrt{x^2 - 36 \ }}{x} \ + \ C_1 \)

\(\displaystyle \ln\bigg|x \ + \ \sqrt{x^2 - 36 \ }\bigg| \ - \ \dfrac{\sqrt{x^2 - 36 \ }}{x} \ + \ C_1 \ - \ ln(6) \)


\(\displaystyle Let \ \ C \ = \ C_1 \ - \ ln(6).\)


\(\displaystyle \boxed{ \ \ \ln\bigg|x \ + \ \sqrt{x^2 - 36 \ }\bigg| \ - \ \dfrac{\sqrt{x^2 - 36 \ }}{x} \ + \ C \ \ } \)