Trig substitution integral

**DBA** · February 26th 2010, 01:58 PM

$<br /> \int \frac{\sqrt{x^{2}-4}}{x^2}dx<br />$

I started with setting

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

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

Then I rewrite it...

$<br /> \int \frac{2tan\theta}{4sec^{2}\theta} * 2 sec \theta tan\theta d\theta = \int \frac{tan^{2}\theta}{sec\theta}d\theta = \int \frac{sin^{2}\theta}{cos\theta}d\theta <br />$

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

**ANDS!** · February 26th 2010, 02:13 PM

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

**DBA** · February 26th 2010, 02:33 PM

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

**ANDS!** · February 26th 2010, 02:56 PM

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

**DBA** · February 26th 2010, 03:38 PM

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

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

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

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

**Keithfert488** · February 26th 2010, 09:20 PM

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

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