# Integrate $\int \frac{\sqrt{9x^2-1}}{2x}dx$

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?

• The title has $16$. The body of the q has $1$. Which is it? – Oscar Lanzi Mar 13 '19 at 13:47
• My apologies, it should have been $1$. – Eldar Rahimli Mar 13 '19 at 13:49
• – lab bhattacharjee Mar 16 '19 at 9:16

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?
• 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? – Eldar Rahimli Mar 13 '19 at 13:49
• @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. – J.G. Mar 13 '19 at 13:51
• 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 $\frac34$. – José Carlos Santos Mar 13 '19 at 13:52
Hint Apply the subsititon $$x= \dfrac{\sec \theta}{3}$$ to simplify the integral .