# For which values a, b does this integral converge

Determine all $$a$$ and $$b$$ for which the this converges:

$$\int_0^{\pi/2}\frac{\mathrm{d}x}{(\sin^ax^2)(\cos^bx)}$$

I'm not sure how to approach the problem. I've tried using the comparison test, setting:

$$f(x)=1/(sin^ax^2)(cos^bx)$$ $$g(x)=1/(sin^ax^2)$$ then doing $$\lim_{x \to 0}f(x)/g(x)dx = 1/(cos^bx) = 1$$

then by comparison test I should get that whenever g(x) converges, so does f(x) -> but I'm not sure where to continue from here

Ideally I would like to show that g(x) converges whenever a = ??? and then I can also solve for b

Some help or a general direction would be appreciated note that I saw this question: For which values of $p$ and $q$ does an improper integral converge However this one is different, and I did not understand how to apply the answers given there

$$f(a, b) =\int_0^{\pi/2}\frac{\mathrm{d}x}{(\sin^ax^2)(\cos^bx)}$$.
Around $$x=0$$, $$\dfrac1{\sin^ax^2} \approx \dfrac1{x^{2a}} =x^{-2a}$$ and $$\dfrac1{\cos^bx} \approx 1$$, so the integral is like $$\lim_{r \to 0}\int_r^c x^{-2a}dx =\lim_{r \to 0}\dfrac{x^{-2a+1}}{-2a+1}|_r^c$$ and this diverges for $$-2a+1 < 0$$ or $$a > \frac12$$.
Similarly, around $$x = \pi/2$$, $$\cos(\pi/2-x) =\sin(x) \approx x$$ so so the integral is like $$\lim_{r \to 0}\int_r^c x^{-b}dx =\lim_{r \to 0}\dfrac{x^{-b+1}}{-b+1}|_r^c$$ and this diverges for $$b > 1$$.
You can show that the integral converges for $$a < \frac12$$ and $$b < 1$$.
For $$a=\frac12$$ and $$b = 1$$ the integral behaves like $$\lim_{r \to 0}\int_r^c \dfrac{dt}{t} =\lim_{r \to 0}\ln(t)|_r^c$$ and this diverges.