Evaluate $\mathbf{\int }{\mathbf{e}}^{\mathbf{(}\mathbf{a}\mathbf{+}\mathbf{ib}\mathbf{)}\mathbf{x}}\mathbf{dx}$and take real and imaginary parts to show that:

$\mathbf{\int }{\mathbf{e}}^{\mathbf{ax}}\mathbf{cos}\mathbf{}\mathbf{b}\mathbf{}\mathbf{xdx}\mathbf{=}\frac{{\mathbf{e}}^{\mathbf{ax}}\mathbf{(}\mathbf{a}\mathbf{}\mathbf{cos}\mathbf{}\mathbf{b}\mathbf{}\mathbf{x}\mathbf{+}\mathbf{b}\mathbf{}\mathbf{sin}\mathbf{}\mathbf{b}\mathbf{}\mathbf{x}\mathbf{)}}{{\mathbf{a}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}}$

The given expression is$\int {\mathrm{e}}^{(\mathrm{a}+\mathrm{ib})\mathrm{x}}\mathrm{dx}$ .

Representing the complex number in rectangular form means writing the given complex number in the form of $\mathbf{x}\mathbf{+}\mathbf{iy}$, in which is the real part and y is the imaginary part.

The given question is $\int e^{(a+ib)x}dx$.

role="math" localid="1658833021638" $$\begin{gathered} \int e^{(a+ib)x}dx=\frac{e^{{}^{(a+ib)x}}}{a+ib} \\ \int e^{(a+ib)x}dx=\frac{e^{ax}\cdot e^{ibx}}{a+ib} \\ \int e^{(a+ib)x}dx=\frac{e^{ax}[\cos bx+i\sin bx]}{a+ib} \end{gathered}$$

Multiply the numerator and the denominator with the complex conjugate.

$$\begin{gathered} \int {\mathrm{e}}^{(\mathrm{a}+\mathrm{ib})\mathrm{x}}\mathrm{dx}=\frac{{\mathrm{e}}^{\mathrm{ax}}\left[\mathrm{a}\cos \mathrm{b}\mathrm{x}+\mathrm{i}\sin \mathrm{b}\mathrm{x}\right]}{{\mathrm{a}}^2+{\mathrm{b}}^2}\times \frac{\mathrm{a}-\mathrm{ib}}{\mathrm{a}-\mathrm{ib}} \\ \int {\mathrm{e}}^{(\mathrm{a}+\mathrm{ib})\mathrm{x}}\mathrm{dx}=\frac{{\mathrm{e}}^{\mathrm{ax}}\left[\mathrm{a}\cos \mathrm{b}\mathrm{x}+\mathrm{ai}\sin \mathrm{b}\mathrm{x}-\mathrm{ib}\cos \mathrm{bx}+\mathrm{b}\sin \mathrm{b}\mathrm{x}\right]}{{\mathrm{a}}^2+{\mathrm{b}}^2} \\ \int {\mathrm{e}}^{(\mathrm{a}+\mathrm{ib})\mathrm{x}}\mathrm{dx}=\frac{{\mathrm{e}}^{\mathrm{ax}}\left(\mathrm{a}\cos \mathrm{b}\mathrm{x}+\mathrm{i}\sin \mathrm{b}\mathrm{x}\right)+{\mathrm{ie}}^{\mathrm{ax}}\left(\mathrm{a}\sin \mathrm{b}\mathrm{x}-\mathrm{b}\cos \mathrm{bx}\right)}{{\mathrm{a}}^2+{\mathrm{b}}^2} \end{gathered}$$........(1)

$$\begin{gathered} \int {\mathrm{e}}^{(\mathrm{a}+\mathrm{ib})\mathrm{x}}\mathrm{dx}=\int {\mathrm{e}}^{\mathrm{ax}}\cdot {\mathrm{e}}^{\mathrm{ibx}}\mathrm{dx} \\ \int {\mathrm{e}}^{(\mathrm{a}+\mathrm{ib})\mathrm{x}}\mathrm{dx}=\int {\mathrm{e}}^{\mathrm{ax}}(\cos \mathrm{b}\mathrm{x}+\mathrm{i}\sin \mathrm{b}\mathrm{x})\mathrm{dx} \\ \int {\mathrm{e}}^{(\mathrm{a}+\mathrm{ib})\mathrm{x}}\mathrm{dx}=\int {\mathrm{e}}^{\mathrm{ax}}\cos \mathrm{b}\mathrm{x}\mathrm{dx}+\mathrm{i}\int {\mathrm{e}}^{\mathrm{ax}}\sin \mathrm{b}\mathrm{x}\mathrm{dx}.......\left(2\right) \end{gathered}$$

Using equation (1) and (2).

$\int {\mathrm{e}}^{\mathrm{ax}}\cos \mathrm{b}\mathrm{x}\mathrm{dx}+\mathrm{i}\int {\mathrm{e}}^{\mathrm{ax}}\sin \mathrm{b}\mathrm{x}\mathrm{dx}=\frac{{\mathrm{e}}^{\mathrm{ax}}(\mathrm{a}\cos \mathrm{b}\mathrm{x}+\mathrm{b}\sin \mathrm{b}\mathrm{x})+{\mathrm{ie}}^{\mathrm{ax}}(\mathrm{a}\sin \mathrm{b}\mathrm{x}-\mathrm{b}\cos \mathrm{b}\mathrm{x})}{{\mathrm{a}}^2+{\mathrm{b}}^2}$

Equate the real part on both sides.

$\int e^{ax}cosbxdx=\frac{e^{ax}(a\cos bx+b\sin bx)}{a^2+b^2}$

Therefore, the evaluation of $\int {\mathrm{e}}^{(\mathrm{a}+\mathrm{ib})\mathrm{x}}\mathrm{dx}=\frac{{\mathrm{e}}^{\mathrm{ax}}(\mathrm{a}\cos \mathrm{b}\mathrm{x}+\mathrm{b}\sin \mathrm{b}\mathrm{x})+{\mathrm{ie}}^{\mathrm{ax}}\left(\mathrm{a}\sin \mathrm{b}\mathrm{x}-\mathrm{b}\cos \mathrm{bx})\right)}{{\mathrm{a}}^2+{\mathrm{b}}^2}$and the function$\int {\mathrm{e}}^{\mathrm{ax}}\cos \mathrm{b}\mathrm{xdx}=\frac{{\mathrm{e}}^{\mathrm{ax}}(\mathrm{a}\cos \mathrm{b}\mathrm{x}+\mathrm{b}\sin \mathrm{b}\mathrm{x})}{{\mathrm{a}}^2+{\mathrm{b}}^2}$ has been showed.