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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?

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

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

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

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    $\begingroup$ 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. $\endgroup$ – J.G. Sep 15 at 20:43

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