Yes it is possible to do using Integration by Parts, but it is a LOT of work!

First you will need to know what $\displaystyle \begin{align*} \int{ \mathrm{e}^{2\,x}\sin{(x)}\,\mathrm{d}x} \end{align*}$ is...

$\displaystyle \begin{align*} I &= \int{\mathrm{e}^{2\,x}\sin{(x)}\,\mathrm{d}x} \\ I &= \frac{1}{2}\,\mathrm{e}^{2\,x} \sin{(x)} - \int{ \frac{1}{2}\,\mathrm{e}^{2\,x}\cos{(x)}\,\mathrm{d}x} \\ I &= \frac{1}{2}\,\mathrm{e}^{2\,x} \sin{(x)} - \frac{1}{2} \int{ \mathrm{e}^{2\,x}\cos{(x)}\,\mathrm{d}x } \\ I &= \frac{1}{2} \,\mathrm{e}^{2\,x}\sin{(x)} - \frac{1}{2}\, \left[ \frac{1}{2}\,\mathrm{e}^{2\,x} \cos{(x)} - \int{ -\frac{1}{2} \,\mathrm{e}^{2\,x}\sin{(x)}\,\mathrm{d}x} \right] \\ I &= \frac{1}{2} \,\mathrm{e}^{2\,x} \sin{(x)} - \frac{1}{4}\,\mathrm{e}^{2\,x} \cos{(x)} - \frac{1}{4}\, I \\ \frac{5}{4}\,I &= \frac{1}{2}\,\mathrm{e}^{2\,x}\sin{(x)} - \frac{1}{4}\,\mathrm{e}^{2\,x}\cos{(x)} \\ I &= \frac{2}{5}\,\mathrm{e}^{2\,x} \sin{(x)} - \frac{1}{5}\,\mathrm{e}^{2\,x} \cos{(x)} \end{align*}$

So that means $\displaystyle \begin{align*} \int{ \mathrm{e}^{2\,x} \sin{(x)}\,\mathrm{d}x} = \frac{2}{5}\,\mathrm{e}^{2\,x} \sin{(x)} - \frac{1}{5}\,\mathrm{e}^{2\,x} \cos{(x)} + C \end{align*}$. Now applying it to your problem $\displaystyle \begin{align*} \int{ -x\,\mathrm{e}^{2\,x} \sin{(x)}\,\mathrm{d}x} \end{align*}$ you need to use integration by parts with $\displaystyle \begin{align*} u = -x \implies \mathrm{d}u = -\mathrm{d}x \end{align*}$ and $\displaystyle \begin{align*} \mathrm{d}v = \mathrm{e}^{2\,x} \sin{(x)} \,\mathrm{d}x \implies v = \frac{2}{5}\mathrm{e}^{2\,x} \sin{(x)} - \frac{1}{5}\,\mathrm{e}^{2\,x}\cos{(x)} \end{align*}$. Have a go...