Answer
$$\frac{{{e^{ax}}}}{{{a^2} + {b^2}}}\left( {a\sin bx - b\cos bx} \right) + C$$
Work Step by Step
$\begin{gathered}
\int {{e^{ax}}\sin bx} dx \hfill \\
{\text{Use tabular integration to evaluate the integral}} \hfill \\
\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}
{{e^{ax}}{\text{ and its derivatives}}}&{}&{\sin bx{\text{ and its integrals}}} \\
{{e^{ax}}}&{\mathop {\left( + \right)}\limits_ \searrow }&{\sin bx} \\
{a{e^{ax}}}&{\mathop {\left( - \right)}\limits_ \searrow }&{ - \frac{1}{b}\cos bx} \\
{{a^2}{e^{ax}}}&{\left( + \right)}&{ - \frac{1}{{{b^2}}}\sin bx}
\end{array} \hfill \\
\end{gathered} $ $$\eqalign{
& {\text{We need stop here because it is the same as the first row except }} \cr
& {\text{for constants }}\left( {{a^2}{\text{ on the left and }} - \frac{1}{{{b^2}}}{\text{ on the right}}} \right).{\text{ Then}}{\text{:}} \cr
& {\text{We interpret the table as saying }} \cr
& \int {{e^{ax}}\sin bx} dx = \left( {{e^{ax}}} \right)\left( { - \frac{1}{b}\cos bx} \right) - \left( {a{e^{ax}}} \right)\left( { - \frac{1}{{{b^2}}}\sin bx} \right) + \int {\left( {{a^2}{e^{ax}}} \right)\left( { - \frac{1}{{{b^2}}}\sin bx} \right)} dx \cr
& \,\int {{e^{ax}}\sin bx} dx = - \frac{1}{b}{e^{ax}}\cos bx + \frac{{a{e^{ax}}}}{{{b^2}}}\sin bx - \frac{{{a^2}}}{{{b^2}}}\int {{e^{ax}}\sin bx} dx \cr
& \cr
& {\text{Solve for }}\int {{e^{ax}}\sin bx} dx \cr
& \,\int {{e^{ax}}\sin bx} dx + \frac{{{a^2}}}{{{b^2}}}\int {{e^{ax}}\sin bx} dx = - \frac{1}{b}{e^{ax}}\cos bx + \frac{{a{e^{ax}}}}{{{b^2}}}\sin bx + C \cr
& \,\left( {1 + \frac{{{a^2}}}{{{b^2}}}} \right)\int {{e^{ax}}\sin bx} dx = - \frac{1}{b}{e^{ax}}\cos bx + \frac{{a{e^{ax}}}}{{{b^2}}}\sin bx + C \cr
& \,\int {{e^{ax}}\sin bx} dx = \left( {\frac{{{b^2}}}{{{a^2} + {b^2}}}} \right)\left( { - \frac{1}{b}{e^{ax}}\cos bx + \frac{{a{e^{ax}}}}{{{b^2}}}\sin bx} \right) + C \cr
& \,\int {{e^{ax}}\sin bx} dx = \left( {\frac{1}{{{a^2} + {b^2}}}} \right)\left( { - b{e^{ax}}\cos bx + a{e^{ax}}\sin bx} \right) + C \cr
& \,\int {{e^{ax}}\sin bx} dx = \frac{{{e^{ax}}}}{{{a^2} + {b^2}}}\left( {a\sin bx - b\cos bx} \right) + C \cr} $$
