2
$\begingroup$

Calculate $$\int \sqrt{x^2-4}\,dx$$

I tried using substitution for $x = 2\sec\theta $ but I got stuck at $4\int \tan^2\theta \sec\theta d\theta$

$\endgroup$
  • $\begingroup$ Are you sure it should be a four factored out? $\endgroup$ – Quality Mar 23 '15 at 19:24
  • $\begingroup$ @LearningMath The 4 looks right to me. 2 from the square root and 2 from dx. $\endgroup$ – Mike Mar 23 '15 at 19:33
3
$\begingroup$

Write $$ \int \tan^2 \theta \sec \theta \, d\theta = \int \tan \theta (\sec \theta \tan \theta) \, d\theta$$ and integrate by parts. You should (after some algebra) be able to express the integral using the integral of the secant.

$\endgroup$
1
$\begingroup$

$$\int tan^2 \theta \sec \theta d\theta = \int \sec^3 \theta d\theta - \int \sec \theta d\theta$$

Now, just follow: http://en.wikipedia.org/wiki/Integral_of_secant_cubed

$\endgroup$
0
$\begingroup$

$$\int{\tan^2\theta \sec\theta} \ d\theta = \int{\frac{\sin^2\theta}{\cos^3\theta}}d\theta = \int{\frac{d\theta}{\cos^3\theta}} - \int{\frac{d\theta}{\cos\theta}}$$

For both integrals, multiply the numerator and denominator by $\cos\theta$ and use the change of variable $t = \sin\theta$.

Things should be easy from here.

$\endgroup$
0
$\begingroup$

Alternatively, one may recall that $$ \cosh^2 u -\sinh^2 u=1 $$ or $$ \cosh^2 u -1=\sinh^2 u $$ then you may try the change of variable $\displaystyle x=2\cosh u$, giving $\displaystyle dx=2\sinh u\: du\,$ and $$ \int\sqrt{x^2-4}\:dx =4\int\sinh^2 u \:du $$ Hoping you can take it from here.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.