How do I solve the following trigonometric substitution? I don't understand why the answer is arcsin

Question

So, I've been doing this problem over and over for a while. I want to get it right and I figure I've almost got it. However, I don't understand why the answer is what it is.

Anyway this is the integral, and its answer:

$$ \int \frac{\sqrt{x^2 - 9}}{x^3} dx = \frac{\sqrt{x^2-9}}{2x^2} - \sin^{-1}{\frac{x}{3}} + C$$

I don't understand where the first rational came from, or the inverse sine. I have an idea of how those things work, but I just don't know how the solution led to that. Anyway, this is what I tried:

Trigonometric substitution:

$$ x = 3\sec{u} \\ dx = 3\sec{u}\tan{u} du $$

Anyway, this integral is long, I ended up at this point:

$$\frac{1}{6}[u - \sin{u}\cos{u}] + c $$

I think it is correct at that point, but I'm not sure. I'm not sure why the answer ends up looking like it does above. I was wondering if someone could help me out.

Answer 1

$$I= \int \frac{\sqrt{x^2 - 9}}{x^3}\ dx$$

Let $x=\sec u$ as you did

$dx=\sec u \tan u du$

$$I= \int \frac{3\tan^2 u}{\sec^3u}\ du $$

$$\frac{I}{3}= \int \cos u\ du -\int \cos^3u\ du $$

Can you finish?

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As OP ended up with $$\frac{1}{6}(u - \sin{u}\cos{u}) + C $$ where $u=\sec^{-1}(x)$

Construct a triangle:

$$\sin u=\frac{\sqrt{x^2-1}}{x} \implies u=\sin^{-1}(\frac{\sqrt{x^2-1}}{x}) $$

You can manipulate it to any inverse function you want :)

Answer 2

We have $x\geq3$ or $x\leq-3$.

Consider two cases.

1. $x=\frac{3}{\cos{u}},$ where $u\in\left[0,\frac{\pi}{2}\right).$

Thus, $$\int\frac{\sqrt{x^2-9}}{x^3}dx=\int\frac{3\tan{u}}{\frac{27}{\cos^3u}}\cdot\frac{3\sin{u}}{\cos^2u}du=\frac{1}{3}\int\sin^2udu=$$ $$=\frac{1}{6}\int(1-\cos2u)du=\frac{1}{6}(u-\frac{1}{2}\sin2u)+C=$$ $$=\frac{1}{6}\left(\arccos\frac{3}{x}-\sqrt{1-\frac{9}{x^2}}\cdot\frac{3}{x}\right)+C=\frac{1}{6}\left(\arccos\frac{3}{x}-\frac{3}{x^2}\sqrt{x^2-9}\right)+C.$$ 2. $x=-\frac{3}{\cos{u}},$ where $u\in\left[0,\frac{\pi}{2}\right).$

Thus, $$\int\frac{\sqrt{x^2-9}}{x^3}dx=\int\frac{3\tan{u}}{-\frac{27}{\cos^3u}}\cdot\left(-\frac{3\sin{u}}{\cos^2u}\right)du=\frac{1}{3}\int\sin^2udu=$$ $$=\frac{1}{6}\int(1-\cos2u)du=\frac{1}{6}(u-\frac{1}{2}\sin2u)+C=$$ $$=\frac{1}{6}\left(\arccos\frac{3}{-x}-\sqrt{1-\frac{9}{x^2}}\cdot\frac{3}{-x}\right)+C=\frac{1}{6}\left(\pi-\arccos\frac{3}{x}-\frac{3}{x^2}\sqrt{x^2-9}\right)+C=$$ $$=\frac{1}{6}\left(-\arccos\frac{3}{x}-\frac{3}{x^2}\sqrt{x^2-9}\right)+C.$$