# How to solve $\int \frac{\sqrt{x^2-1}}{x^2}dx$ with trig substitution?

I can't find a good trig identity to use on this equation, and I'm stuck. I feel like there's something based on $$1 + \tan^2(x) = \sec^2(x)$$ that will let me solve it, but I can't figure out what.

$$\int \frac{\sqrt{x^2-1}}{x^2}dx = A$$

Let $$x = \sec \theta$$

Then $$dx = \sec \theta \tan \theta$$

$$A =\int \frac{\sqrt{\sec^2 \theta - 1}}{\sec^2\theta}\sec\theta\tan\theta \ d\theta$$

$$= \int \frac{\sqrt{(\tan^2\theta +1 ) - 1}}{\sec^2\theta}\sec\theta\tan\theta \ d\theta$$

$$=\int \frac{\tan^2\theta}{\sec\theta}d \theta$$

How can I solve this integral with trigonometric substitution?

hint: $$\dfrac{\tan^2\theta}{\sec \theta} = \sec \theta - \cos \theta$$

• Thank you, that really helps! Could you show your steps inbetween? If I know how it works, it's easier to memorize. – LuminousNutria Sep 15 at 19:32
• @LuminousNutria: $\dfrac{\tan^2}{\sec} = \dfrac{\sec^2-1}{\sec} = \sec - \dfrac{1}{\sec} = \sec - \cos$. – DeepSea Sep 15 at 19:34
• Awesome! thank you – LuminousNutria Sep 15 at 19:37

It is actually more convenient to integrate directly,

$$I=\int \frac{\sqrt{x^2-1}}{x^2}dx=-\int \sqrt{x^2-1}d\left(\frac{1}{x}\right)$$

Integrate-by-parts,

$$I=-\frac{1}{x}\sqrt{x^2-1} +\int \frac{dx}{\sqrt{x^2-1}}$$ $$=-\frac{1}{x}\sqrt{x^2-1}+\ln\left(\sqrt{x^2-1}+x\right) +C$$

• It depends on how convenient you think it is to evaluate $\int\frac{dx}{\sqrt{x^2-1}}$; some may not spot the substitution $u=\sqrt{x^2-1}+x$. At that point, it would help to go with $x=\sec\theta$ after all. – J.G. Sep 15 at 20:43