Applying integration by part on the first expression gives

$$\begin{align}&\int e^{ax}\cos(bx)d{x} \nonumber\\&= \frac{1}{a}\int\cos(bx)d(e^{ax}) \nonumber\\&=\frac{1}{a}e^{ax}\cos(bx)-\frac{1}{a}\int e^{ax}d(cos(bx)) \nonumber\\&=\frac{1}{a}e^{ax}\cos(bx)+\frac{b}{a}\int e^{ax}\sin(bx)d{x}\label{eq_ecos}\end{align}$$

Similarly, we can have

$$\int e^{ax}\sin(bx)d{x}=\frac{1}{a}e^{ax}\sin(bx)-\frac{b}{a}\int e^{ax}\cos(bx)d{x}$$

Now, solving the two equations will find the final results as

$$\left\{\begin{array}{ll}\int e^{ax}\cos(bx)dx &=\boxed{\frac{e^{ax}}{a^2+b^2}\left(b\sin(bx)+a\cos(ax)\right)+C}\\\int e^{ax}\sin(bx)dx &=\boxed{\frac{e^{ax}}{a^2+b^2}\left(a\sin(bx)-b\cos(ax)\right)+C}\\\end{array}\right.$$