Integrating $\int\frac{\sqrt{16x^2-9}}x\,\mathrm{d}x$?

Question

I am trying to differentiate from my previous question, but I am having trouble in the finishing steps. I have the integral $\int\dfrac{\sqrt{16x^2-9}}x\,\mathrm{d}x$. $$v=4x\implies\mathrm{d}v=4\,\mathrm{d}x$$

$$\int\frac{\sqrt{v^2-9}}{v}\,\mathrm{d}v\implies a=3,\quad v=3\sec\theta,\quad\mathrm{d}v=3\sec\theta\tan\theta\,\mathrm{d}\theta.$$ $$v^2-9=9\sec^2\theta-9=9\left(\sec^2\theta-1\right)=9\tan^2\theta.$$ $$\int\frac{3\tan\theta\cdot3\sec\theta\tan\theta}{3\sec\theta}\,\mathrm{d}\theta=3\int\tan^2\theta\,\mathrm{d}\theta.$$ $$3\int\left(\sec^2\theta-1\right)\mathrm{d}\theta=3\left(\tan\theta-\theta\right).$$

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 $\dfrac{\sqrt{16x^2-9}}3$ and for $\theta$ I got $\sec^{-1}\left(\dfrac{v}3\right)$ or $\cos\left(\dfrac{4x}3\right)$. 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.

Answer 1

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.

Answer 2

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.

Answer 3

I would rather like to try to tackle it with rationalization other than substitution. $$ \begin{aligned} \int \frac{\sqrt{16 x^{2}-9}}{x} d x &=\int \frac{16 x^{2}-9}{x \sqrt{16 x^{2}-9}} d x \\ &=\frac{1}{16} \int \frac{16 x^{2}-9}{x^{2}} d\left(16 x^{2}-9\right) \\ &=\frac{1}{16} \int\left(16-\frac{9}{x^{2}}\right) d\left(\sqrt{16 x^{2}-9}\right) \\ &=\sqrt{16 x^{2}-9}-3 \arctan \left(\frac{\sqrt{16 x^{2}-9}}{3}\right)+C \\ & (\textrm{ OR }\sqrt{16 x^{2}-9}-3 \arccos \left(\frac{3}{4 x}\right)+C) \end{aligned} $$