Convergence of $\int_0^1 \frac{\sin\big(\frac1{x^2}\big)}{\frac\pi2 -\ \rm{arccotg}\ x}$ 
Problem: does the integral $\displaystyle\int_0^1 \frac{\sin\big(\frac1{x^2}\big)}{\frac\pi2 -\ \rm{arccotg}\ x}$ converge?

I know that the function is continuous on $(0,1]$, that the problem is at $x = 0$ and that I'll have to use Abel's (or Dirichlet's) convergence criteria. But I cannot find the functions $g, h$ so that $f = gh$, $g$ is monotonic and $f$ is "known" (in the sense that antiderivative is known or $\int_0^1 f$ converges.)
Could anyone provide a hint? 
Edit: One of my previous attempts was to write
$$
f(x) = (-2)\frac{\sin{\frac1{x^2}}}{x^3}\frac{x^3}{-2\big(\frac\pi2 - {\rm arccotg}\ x\big)}
$$
Then I can prove that the first factor converges (using $t = \frac1{x^2}$). The problem is that I do not know how the prove the monotonicity of the second factor.
 A: First of all, note that $\frac \pi2 - \operatorname{arc ctg} x = \operatorname{arc tg} x$ for $x\geq 0$. If anything, this makes the problem shorter to write, and maybe easier to think of. Also, bear in mind that $\operatorname{arc tg} x\approx x$ near $0$, more precisely:
$$
\lim_{x\to 0 } \frac{\operatorname{arc tg} x}x = 1
\quad \text{and} \quad
\frac 12 x \stackrel{(\star)}< \operatorname{arc tg} x < x \quad \text{for } x\in(0, \varepsilon)
$$
(in fact at least for $x\in(0, 2\pi/3)$ and actually for $x\in (0,0.7421\dots\pi)$; if you don't know these, see the remark at the end).
Now, if we write the integrand as you suggest (and using my little note, and simplifying a bit):
$$
\frac{\sin{\frac1{x^2}}}{x^3}\frac{x^3}{\operatorname{arc tg} x}
$$
and you're asking about the monotonicity of the second factor, then we should try the derivative. I'll leave the calculation as an exercise – let's examine the result:
$$
\big(\frac{x^3}{\operatorname{arc tg} x}\big)' = \frac{x^2}{(\operatorname{arc tg}x)^2} \bigg(3 \operatorname{arc tg} x - \frac x{x^2+1}\bigg).
$$
Because of the inequalities: $(\star)$ and $x^2 + 1 > 1$:
$$
3 \operatorname{arc tg} x - \frac x{x^2+1} > \frac 32 x -  x = \frac 12 x > 0
$$
for $x\in(0,\varepsilon)$, so the second factor is increasing in $(0,\varepsilon)$ (in fact, in $(0, \infty)$), which should yield the convergence of the integral $\displaystyle\int_0^\varepsilon (\dots)$.
Remark. The limit follows from the limit $\frac{\operatorname{tg} x}x\to 1$ and further from $\frac{\sin x}x\to 1$, which is classic, I think. The inequality $\operatorname{arc tg}x < x$ follows from $\operatorname{tg}x > x$, which I suppose is also classic. Finally, given the limit, it must be $\operatorname{arc tg}x > \frac 12 x$ close to zero: for some $x\in(0,\varepsilon)$. The coefficient $\frac 12$ isn't special: anything in $(\frac 13, 1)$ will do fine here, but $\frac 12$ came first to my mind.
