The above integral has a pole on the real axis. The pole is of order $3$. If it was a simple pole, we could use an indented path and find the integral. However, I don’t know of any method if it is a multiple pole. I have referred to many books, and they dealt only with simple poles. Can anyone help me with this problem with multiple poles?
The question was asked in an exam. Was the question wrong, or there is a way to solve this problem?
There is a sense in which this integral regularizes to $-2\pi$ which is option $(a)$. We define for $t\in\Bbb R$: $$ I(t)=PV\int_\Bbb R\frac{\sin 2x}{x-t}\, \mathrm dx=\pi\cos 2t, $$ which exists in the sense of the Cauchy principal value. The function $I(t)$ now acts as a generating function by defining the regularization for $n\in\Bbb N$ $$ \mathcal P\int_\Bbb R\frac{\sin 2x}{x^n}\, \mathrm dx:=\frac{1}{(n-1)!}\partial_t^{n-1}I(t)\Big|_{t=0}. $$ In particular, $$ \mathcal P\int_\Bbb R\frac{\sin 2x}{x^3}\, \mathrm dx=\frac{\pi}{2}\partial_t^2\cos 2t\Big|_{t=0}=-2\pi. $$ This is consistent with Hadamard regularization. For more information see A Generalization of the Cauchy Principal Value by Charles Fox.
Edit:
The regularized value $-2\pi$ can also be arrived at by other means. Since $\sin$ is analytic in the upper and lower halves of the complex plain we may employ the Analytic Principal Value to regularize your integral. To derive the APV for your integral we simply integrate along the two paths $\gamma^\pm$ as shown in the picture and take their average.
In this specific example, the integral along $\gamma^-$ is equal to the integral along $\gamma^+$ so we simply need to evaluate the integral along $\gamma^+$ to find the regularized value. We have $$ \mathcal{APV}\int_\Bbb R\frac{\sin 2x}{x^3}\, \mathrm dx =\int_{\Bbb R\setminus [-1,1]}\frac{\sin 2x}{x^3}\, \mathrm dx+i\int_{\pi}^0\sin(2e^{i\varphi})e^{-2i\varphi}\,\mathrm d\varphi $$ with $$ \int_{\Bbb R\setminus [-1,1]}\frac{\sin 2x}{x^3}\, \mathrm dx=4 \operatorname{Si}(2)-2 \pi +\sin 2+2 \cos 2, $$ and $$ i\int_{\pi}^0\sin(2e^{i\varphi})e^{-2i\varphi}\,\mathrm d\varphi=-4 \operatorname{Si}(2)-\sin 2-2\cos 2, $$ where $\operatorname{Si}(z):=\int_0^z\sin t/t\,\mathrm dt$ is the sine integral. Bringing both results together gives $$ \mathcal{APV}\int_\Bbb R\frac{\sin 2x}{x^3}\, \mathrm dx=-2\pi, $$ which is again option $(a)$.