Is integration of $x\operatorname{cosec}(x)$ defined? Is integration of $x\operatorname{cosec}(x)$ possible? If yes, then what is its closed form; if not, then why is it non-integrable ?

Yes, $x\,\mathrm{cosec}\,x$ is integrable and has a closed form. 
Hint:
Using the subsitution $u = \tan \left(\frac{x}{2}\right)$, we get:
$$\int \frac{1}{\sin x} \, \mathrm{d}x = \ln \left(\tan \left(\frac{x}{2}\right)\right) + \mathrm{C}$$
Combine this result with the use of integration by parts, specifically, let $u= x \implies \mathrm{d}u = \mathrm{d}x$ and $\mathrm{d}v = \frac{1}{\sin x}$.
Using integration by parts with the above , we get that $$\int \frac{x}{\sin x} \, \mathrm{d}x = x \ln \left(\tan \left(\frac{x}{2}\right)\right) + \int \ln \left(\tan \left(\frac{x}{2}\right)\right) \, \mathrm{d}x$$
The above evaluates to 
$$i\left(\mathrm{Li}_2(e^{-ix}) - \mathrm{Li}_2 (e^{ix})\right) + x \left(\ln(1-e^{ix}) - \ln(1+e^{ix})\right) + \mathrm{C}.$$
Where $\mathrm{Li}_2$ is the Polylogarithm function.
