Show that $\sin(x^2)$ is integrable around $\infty$. I have to show that $f(x)=\sin(x^2)$ is integrable on $[1, \infty[$. This is French terminology, so "intégrable" specifically means that the integral of $|f|$ exists.
The only method I know is to compare it to functions of the form $\frac{1}{x^\alpha}$, but it's not eventually smaller or larger than any of these. I can't imagine how it could be asymptotically equivalent to anything useful either, seeing as it oscillates like crazy.

Your function is not Lebesgue integrable. As I commented, the absolute value is not integrable. By $u=x^2$ we get $$\int_1^\infty \frac{|\sin u|}{2\sqrt u}du$$
Now, consider $f(u)=\frac{1}{4\sqrt u}$. Our integrand is red, this last function is blue. We know that $$\int_1^\infty \frac{du}{\sqrt u}$$ diverges. From the graph, we can see that $$\int_1^\infty \frac{|\sin u|}{2\sqrt u}du> \int_1^\infty \frac{1}{4\sqrt u}du$$ by comparing the size of the red and green triangles. Thus this integral diverges.
(I guess one can make this rigorous, but I think the argument is clear and convincing.)

ADD The  intersection with the $x$-axis are the points $x=\pi k$, i.e. the zeroes of $\sin x$. Solving for the extrema (i.e. using derivatives) gives $\tan x=2x$. These points behave asymtotically like $x=\frac \pi 2+\pi k$. Finally, the intersection of the blue graph with the red graph is given by $|\sin x|=\dfrac 1 4$, thus I don't see it impossible to estimate accurately the areas of the triangles in question.  
