$\begin{align}  & \int{{{e}^{ax}}\cos bxdx} \\ & \text{Let }u={{e}^{ax}},\text{ }du=a{{e}^{ax}}dx \\ & \text{ }dv=\cos bx,\text{  }v=\frac{\sin bx}{b} \\ & \int{{{e}^{ax}}\cos bxdx}={{e}^{ax}}\left( \frac{\sin bx}{b} \right)-\int{\frac{\sin bx}{b}(a{{e}^{ax}}dx)} \\ & ={{e}^{ax}}\left( \frac{\sin bx}{b} \right)-\frac{a}{b}\int{{{e}^{ax}}\sin bxdx} \\ & \text{Repeating the process again for }\int{{{e}^{ax}}\sin bx} \\ & \int{{{e}^{ax}}\cos bxdx}={{e}^{ax}}\left( \frac{\sin bx}{b} \right)-\frac{a}{b}\left[ {{e}^{ax}}\int{\sin bxdx}-\int{\left( \tfrac{d}{dx}({{e}^{ax}})\int \sin bxdx \right)dx} \right] \\ & \int{{{e}^{ax}}\cos bxdx}={{e}^{ax}}\left( \frac{\sin bx}{b} \right)-\frac{a}{b}\left[ \frac{-{{e}^{ax}}}{b}-\int{a{{e}^{ax}}\left( -\frac{\cos bx}{b} \right)dx} \right] \\ & \int{{{e}^{ax}}\cos bxdx}={{e}^{ax}}\left( \frac{\sin bx}{b} \right)+\frac{a}{b}{{e}^{ax}}\cos bx-\frac{{{a}^{2}}}{{{b}^{2}}}\int{{{e}^{ax}}\cos bxdx} \\ & \int{{{e}^{ax}}\cos bxdx}+\frac{{{a}^{2}}}{{{b}^{2}}}\int{{{e}^{ax}}\cos bxdx}={{e}^{ax}}\left( \frac{\sin bx}{b} \right)+\frac{a}{b}{{e}^{ax}}\cos bx \\ & \left( 1+\frac{{{a}^{2}}}{{{b}^{2}}} \right)\int{{{e}^{ax}}\cos bxdx}=\frac{{{e}^{ax}}\sin bx}{b}+\frac{a{{e}^{ax}}\cos bx}{{{b}^{2}}} \\ & \frac{{{b}^{2}}+{{a}^{2}}}{{{b}^{2}}}\int{{{e}^{ax}}\cos bxdx}=\frac{b{{e}^{ax}}\sin bx+a{{e}^{ax}}\cos bx}{{{b}^{2}}} \\ & \int{{{e}^{ax}}\cos bxdx}=\frac{b{{e}^{ax}}\sin bx+a{{e}^{ax}}\cos bx}{{{b}^{2}}+{{a}^{2}}} \\ & \int{{{e}^{ax}}\cos bxdx}=\frac{1}{{{b}^{2}}+{{a}^{2}}}\left[ b{{e}^{ax}}\sin bx+a{{e}^{ax}}\cos bx \right]+C \\\end{align}$