# Integrating $\int\frac{\sqrt{16x^2-9}}{x}dx$?

I am trying to differentiate from my previous question, but I am having trouble in the finishing steps. I have the integral $\int\frac{\sqrt{16x^2-9}}{x}dx$. $$v=4x \hspace{15pt}dv=4dx$$

$$\int\frac{\sqrt{v^2-9}}{v}dv\rightarrow a=3\hspace{10pt}v=3\sec(\theta)\hspace{10pt}dv=3\sec\theta\tan\theta d\theta$$ $$v^2-9=9\sec^2\theta-9=9(\sec^2\theta-1)=9\tan^2\theta$$ $$\int\frac{3\tan\theta 3\sec\theta\tan\theta}{3\sec\theta}d\theta\rightarrow 3\int \tan^2\theta d\theta$$ $$3\int \sec^2\theta-1d\theta\rightarrow 3\left(\tan\theta-\theta\right)$$

And this where I feel I am going wrong. So from my understanding since I used $\sec\theta$ for the substitution the triangle I am using to use this integral has a hypotenuse of $v$ or $4x$ since I initially assigned that value to b and the adjacent side of $\theta$ is $a$ or $3$, and for the missing side I got $\sqrt{v^2-9}$ or $\sqrt{16x^2-9}$. So for $\tan\theta$ I got $\frac{\sqrt{16x^2-9}}{3}$ and for $\theta$ I got $\sec^{-1}\left(\frac{v}{3}\right)$ or $\cos\left(\frac{4x}{3}\right)$. So my final answer is $$\sqrt{16x^2-9}-3\cos\left(\frac{4x}{3}\right)$$ but this is wrong so if someone could tell me where I am going wrong it'd be greatly appreciated. Thanks in advance.

The $\cos$ term is obviously wrong, because differentiating it will give you a $\sin$ term.

If $v = \sec \theta =1/\cos \theta$, $\theta =\cos^{-1} (1/v)$, where this means inverse cosine, not reciprocal of cosine.

You seems to have confused $\sec^{-1}$ as inverse secant with $1/\cos$. Since either meaning could be correct, you have to be careful.

• Thanks, I was indeed confused on that. – Kenshin Sep 7 '14 at 6:04

You're almost there...

Since $x = \frac {3}{4} \sec \theta$, $\sec \theta = \frac {4x}{3}$; thus $\cos \theta = \frac {3}{4x}$ and $\theta = \arccos \frac {3}{4x}$.

By the Pythagorean theorem, $\tan \theta = \frac {\sqrt {16x^2-9}} {3}$. With a little bit of cancellation, your final answer will be...

$$\sqrt {16x^2-9} - 3 \arccos \frac {3}{4x} + C$$

NOTE: I used $\arccos \theta$ versus $\cos^{-1}\theta$ for clarity.