Find limit of an integral with variable upper bounds. $$\lim_{x\to0+}\frac1x\int\limits_{-x}^x\sqrt[t^2]{\cos t}\,\mathrm dt$$


As I understand  $\displaystyle\int_{-x}^{x} \sqrt[t^2]{\cos(t)}\,\mathrm dt = 2\sqrt[x^2]{\cos(x)}$. But I am not sure if it is right and what to do next.
The integrand is defined over $[-\pi/2,\pi/2]\setminus\{0\}$. However the function has a removable singularity at $0$, because
$$
\lim_{x\to0}(\cos x)^{1/x^2}=e^{-1/2}
$$
which can be proved by considering
$$
\lim_{x\to0}\frac{\log\cos x}{x^2}=\lim_{x\to0}\frac{1}{2}\frac{\log(1-\sin^2x)}{\sin^2x}\frac{\sin^2x}{x^2}=-\frac{1}{2}
$$
If we extend the function, we get a continuous function, let's call it $f$, and we have
$$
\int_{-x}^x (\cos t)^{1/t^2}\,dt=\int_{-x}^x f(t)\,dt
$$

Now you have
$$
\lim_{x\to0}\frac{1}{x}\int_{-x}^x f(t)\,dt
$$
with $f$ continuous in a neighborhood of $0$. Now apply the fundamental theorem of calculus with the advantage that, in this case,
$$
\int_{-x}^x f(t)\,dt=2\int_0^x f(t)\,dt
$$
Using L'Hospital we get


  $$\lim_{x\to0+}\frac1x\int\limits_{-x}^x\sqrt[t^2]{\cos t}\,\mathrm dt =
\lim_{x\to0+}\frac2x\int\limits_{0}^x\sqrt[t^2]{\cos t}\,\mathrm dt = 
\lim_{x\to0+}2\sqrt[x^2]{\cos x}\,\mathrm  = 2e^{-\frac12} $$
  because
  $$\lim_{x\to0+}\frac{ln(cosx)}{x^2} = 
\lim_{x\to0+}\frac{-sinx}{2xcosx} = -\frac12 $$