# Thread: Trig substitution integral

1. ## Trig substitution integral

$\displaystyle \int \frac{\sqrt{x^{2}-4}}{x^2}dx$

I started with setting

$\displaystyle x=2sec \theta$ and $\displaystyle dx=2sec \theta tan \theta$
and $\displaystyle x^{2} = 4sec^{2} \theta$

$\displaystyle \tan^{2} \theta = \sec^{2} \theta -1$

Then I rewrite it...

$\displaystyle \int \frac{2tan\theta}{4sec^{2}\theta} * 2 sec \theta tan\theta d\theta = \int \frac{tan^{2}\theta}{sec\theta}d\theta = \int \frac{sin^{2}\theta}{cos\theta}d\theta$

When I go on I do not get anything what helps me solving it.
Thanks for any help...

2. How many anti-derivatives are you meant to know - because there is an anti-derivative here that you could use.

3. I can not think of any anti-derivative I could use. I am not sure if I understand your question.

4. If you make the substitution, $\displaystyle 1-sin^{2}(\theta)$ into your equation, it will reduce down to $\displaystyle sec(\theta)-cos(\theta)$. Secant isn't a "top of my head" integral, but it is good to know.

5. Sorry but I do not get it.. Where can I make the substitution of
$\displaystyle 1-sin^{2}(\theta) = cos^{2} \theta$

I have
$\displaystyle \int \frac{sin^{2}\theta}{cos\theta}d\theta$

I do not knwo how you get to ...
$\displaystyle sec(\theta)-cos(\theta)$

Even if I use
$\displaystyle 1-cos^{2}(\theta) = sin^{2} \theta$
I don't get a term that helps me...

6. $\displaystyle \frac{\sin^2x}{\cos x}=\frac{1-\cos^2x}{\cos x}=\frac{1}{\cos x}-\frac{\cos^2x}{\cos x}=\sec x-\cos x$