<s>[INST] Integrating $\int\frac{\sqrt{16x^2-9}}x\,\mathrm{d}x$? 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.
 [/INST] 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.
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