I know about elementary functions and their integrals, also some elementary function does not have elementary anti-derivative. For example $\frac{\sin(x)}{x}$ , $x^x$ does not have elementary anti-derivatives in spite of being elementary functions.
The given integral $\int \frac{1}{(x\sin(x))^2}\,dx$ , if I start solving it by putting,
$$\sin^2(x) = (1 - \cos(2x))/2$$
$$\int \frac{1}{x^2(1 - \cos(2x))/2}\,dx$$
$$\int \frac{2}{x^2(1 - \cos(2x))}dx$$
dividing and multiplying by $(1 - \cos(2x))$ , we get
$$2\int \frac{1-\cos(2x)}{x^2(1 - \cos(2x))^2}\,dx$$
replacing numerator $(1 - \cos(2x))$ with $2 \sin^2(x)$
$$2\int \frac{2 \sin^2(x)}{x^2(1 - \cos(2x))^2}\,dx$$
after rearranging the above equation we get:
$$4\int \frac{ \sin^2(x)}{x^2}\frac{1}{(1 - \cos(2x))^2}\,dx$$
$$4\int \left(\frac{ \sin(x)}{x}\right)^{\!2}\frac{1}{(1 - \cos(2x))^2}\,dx$$
Now as the given integral contains $\frac{\sin(x)}{x}$ which has no elementary anti-derivative, is this the sufficient condition to say that the given function cannot be integrated?
If this is so, then can we say that any function containing $\frac{\sin(x)}{x}$ , $x^x$ or any such functions cannot have elementary anti-derivative?
EDIT : Please also provide a solution for this problem as the result is calculated by @Dr.SonnhardGraubner in comments.