trig substitution $\int_{2}^{\infty}\frac{1}{x\sqrt{x^2-4}}dx$ $\int_{2}^{\infty}\frac{1}{x\sqrt{x^2-4}}dx$

I let $x = 2 \sec(\theta)$, then $dx = 2\sec(\theta) \tan(\theta) d \theta$

$$\int \frac{2\sec(\theta)\tan(\theta)}{2\sec(\theta)2\tan(\theta)}d\theta = \frac{1}{2}\theta + C = \frac{1}{2} \sec^{-1}\left(\frac{x}{2}\right) + C$$

Therefore the answer would be this after simplification?:

All I did is:

$$\int_{2}^{\infty} f(x)dx = \int_{2}^{3} f(x)dx + \int_{3}^{\infty} f(x)dx$$

$$\lim_{A\to2^+} \frac{1}{2}\left(\sec^{-1}\left( \frac{x}{2}\right)\right)\bigg|_{A}^{3} + \lim_{B\to\infty} \frac{1}{2}\left(\sec^{-1}\left( \frac{x}{2}\right)\right)\bigg|_{3}^{B}$$

$$= \frac{1}{2} (sec^{-1} (\frac{3}{2}) - 0)+ \frac{1}{2}(\pi/2 - sec^{-1}(3/2)) = \pi/4$$ Converges. 

Is this right?
POINT OUTS: You did not change limits. and it looks weird.
But to solve via your method you just need to compute:
$$\int^\frac{\pi}{2} _0 \frac{d \theta}{2}=\frac{\pi}{4}$$
My method: let $x=\frac{1}{t}$
$$\int^0_\frac{1}{2} \frac{dt}{\sqrt{1-4t^2}}$$
Which is equivalent to $$\int^0 _\frac{1}{2} 
\frac{dt}{\sqrt{1-(2t)^2}}$$

Now i hope you know $$\int \frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}(\frac{x}{a})+C$$
hint
the integral is convergent since

$$f (x)\sim \frac {1}{x^2}\;\;(x\to\infty) $$

$$f(x)\sim \frac{1}{4\sqrt{x-2}}\;\;(x\to2^+)$$

to find its value,
put $$x=2\cosh (t) =2\frac{e^t+e^{-t}}{2}.$$
 $x=2\implies t=0$

$x\to \infty\implies t \to \infty $.

on the other hand,

$$dx=2\sinh (t) $$
and
$$\cosh^2 (t)-\sinh^2 (t)=1.$$

thus

$$x\sqrt {x^2-4}=4\cosh(t)\sinh(t). $$

the integral becomes

$$\int_0^\infty \frac{dt}{e^t+e^{-t}} $$ 
$$\color {green}{\boxed {=[\arctan(e^t)]_0^\infty=\frac {\pi}{4}}}$$