How can i prove that this integral is convergent/divergent This is my equation:

$$\int_0^{\pi/4} \frac{dx}{x\sin2x}$$

I wish to prove that it's convergent or divergent, by $P$ test and/or comparison test, but it does not seem to be applicable...

Is it correct, that $1/x$ is smaller or equal to $\frac {1}{x\sin2x}$?

That would mean, $\frac 1x$ is divergent by $p$ test... since it's divergent by $p$ test, and $\frac {1}{x\sin2x}$ is bigger, $\frac {1}{x\sin2x}$ is divergent as well. 

Is this correct?
Over the interval $I=\left(0,\frac{\pi}{2}\right)$, $0&lt;\sin x&lt;x$, hence for any $\varepsilon\in I$:
$$\int_{\varepsilon}^{\pi/4}\frac{dx}{x\sin(2x)}\geq\int_{\varepsilon}^{\pi/4}\frac{dx}{2x^2}=\frac{1}{2\varepsilon}-\frac{2}{\pi}$$
and your integral is clearly divergent.
You can bring it to $\int \frac{dx}{x\sin x}$ integral with some replacements which doesn't converge.