I'm trying to calculate this integral. I'm totally stuck. Can you give me a hint?
I tried some substitutions such as $t = (x^2 - a^2)^p$ for $p \in \lbrace 1, -1, \frac 12, -\frac12 \rbrace$ and integration by parts, but to no avail.
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1$x=a\cosh{u}$ or $x=a\sec{t}$, with the former being easier to evaluate. – Chappers May 07 '16 at 13:49
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"I tried some substitutions" Such as? – Did May 07 '16 at 13:53
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https://www.symbolab.com/solver/definite-integral-calculator/%5Cint%20%5Cfrac%7Bdx%7D%7B%5Csqrt%7Bx%5E%7B2%7D-a%5E%7B2%7D%7D%7D/?origin=button – Mithlesh Upadhyay May 07 '16 at 14:05
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1Possible duplicate of How to integrate $\int \frac{dx}{\sqrt{ax^2-b}}$ – May 07 '16 at 15:03
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Why did you downvote so much? I didn't know about the calculator, and I didn't see the duplicate. thanks for all answers, anyway! – tuxtuxtux May 08 '16 at 09:34
2 Answers
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Let $x=a\cosh\theta$, $\mathrm dx=a\sinh\theta\ \mathrm d\theta$: $$I=\int\frac{a\sinh\theta\ \mathrm d\theta}{a\sinh\theta}=\theta$$
$$I=\cosh^{-1}\left(\frac xa\right)=\ln\left\{\frac xa+\sqrt{\left(\frac xa\right)^2-1}\right\}+C$$
Let $x=a\sec\theta$, $\mathrm dx=a\sec\theta\tan\theta\ \mathrm d\theta$: $$I=\int\frac{a\sec\theta\tan\theta\ \mathrm d\theta}{a\tan\theta}$$
$$=\int\frac{\sec\theta\tan\theta+\sec^2\theta}{\sec\theta+\tan\theta}\mathrm d\theta$$ $$=\int\frac{\mathrm d(\sec\theta+\tan\theta)}{\sec\theta+\tan\theta}$$ $$=\ln\left|\sec\theta+\tan\theta\right|=\ln\left|\frac xa+\sqrt{\left(\frac xa\right)^2-1}\right|+C$$
Kenny Lau
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If you know $\operatorname{arcosh}'(u)=\dfrac1{\sqrt{u^2-1}}$, you just have to set $u=\dfrac xa,\enspace\mathrm d\mkern 1mu u=\dfrac{\mathrm d\mkern 1mu x}{a}$ to get: \begin{align*}\int\frac{\mathrm d\mkern 1mu x}{\sqrt{x^2-a^2}}&=\int\frac{\mathrm d\mkern 1mu u}{\sqrt{u^2-1}}=\operatorname{arcosh}(u)=\ln(u+\sqrt{u^2-1})\\ &=\operatorname{arcosh}\Bigl(\frac xa\Bigr)=\ln(x+\sqrt{x^2-a^2})+\text{c}^\mathrm t. \end{align*}
Bernard
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