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