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What is $\int \frac{\sqrt{9x^2-1}}{2x}dx$?

I tried to form a triangle with $\cos\theta=\frac{1}{3x}$ and $\sin\theta=\frac{\sqrt{9x^2-1}}{3x}$ to use as substitution. But I can't get rid of all the $x$'s to finally integrate with respect to $\theta$.

How is this problem solved?

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If you do $x=\frac43\sec\theta$ and $\mathrm dx=\frac43\sec\theta\tan\theta\,\mathrm d\theta$, then your integral becomes$$\int\frac{4\tan\theta}{\frac83\sec\theta}\frac43\sec\theta\tan\theta\,\mathrm d\theta=2\int\tan^2\theta\,\mathrm d\theta.$$Can you take it from here?

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  • $\begingroup$ Yes, I can, thanks. I have a question though, can you please explain the thought process of how did you come up with $x=\frac43\sec\theta$ substitution. I mean, what hints were involved in the problem for us to use such substitution? $\endgroup$ – Eldar Rahimli Mar 13 '19 at 13:49
  • $\begingroup$ @ElderRahimil The trick when you see a square root of something minus a constant is to try to express the surd in the form $k\sqrt{\sec^2\theta-1]=k|\tan\theta|$. You make whatever substitution that mandates. $\endgroup$ – J.G. Mar 13 '19 at 13:51
  • $\begingroup$ The question has been modified as there had been a typo. Please recheck and correct accordingly. $\endgroup$ – Oscar Lanzi Mar 13 '19 at 13:51
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    $\begingroup$ If we simply had $\sqrt{x^2-1}$, then I would have thought about using $x=\sec\theta$ (since $\sqrt{\sec^2\theta-1}=\tan\theta$). But hat we had here was$$\sqrt{9x^2-16}=4\sqrt{\left(\frac34x\right)^2-1}.$$So, I've added the $\frac43$ factor to compensate for that $\frac34$. $\endgroup$ – José Carlos Santos Mar 13 '19 at 13:52
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Hint Apply the subsititon $x= \dfrac{\sec \theta}{3}$ to simplify the integral .

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