. ${e^{2x}}\sin x$

Let$I = \int {{e^{2x}}\sin x} dx$
$= {e^{2x}}\int {\sin x\,dx} - \int {\left( {\cfrac{d}{{dx}}\left( {{e^{2x}}} \right) \cdot \int {\sin x} \,dx} \right)} dx$

$= {e^{2x}}\left( { - \cos \,x} \right) - \int {2{e^{2x}}\left( { - \cos \,x} \right)dx + } {C_1}$

$= - {e^{2x}}\cos x + 2\int {{e^{2x}}\cos x\,dx + } {C_1}$

$= - {e^{2x}}\cos x + 2\left[ {{e^{2x}}\int {\cos x} dx - \int {\left( {\cfrac{d}{{dx}}\left( {{e^{2x}}} \right) \cdot \int {\cos x} dx} \right)dx} } \right] + {C_1}$

$= - {e^{2x}}\cos x + 2{e^{2x}}\sin x - 4\int {{e^{2x}}\sin x} dx + {C_1}$

$= {e^{2x}}\left( {2\sin x - \cos x} \right) - 4I + {C_1}$

$\therefore$ $5I = {e^{2x}}\left( {2\sin x - \cos x} \right) + {C_1}$
$\Rightarrow$ $I = \cfrac{{{e^{2x}}}}{5}\left( {2\sin x - \cos x} \right) + C$ $\left[ {C = \cfrac{{{C_1}}}{5}} \right]$