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

$\mathbf{\int }{\mathit{e}}^{\mathbf{a}\mathbf{x}}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{b}\mathit{x}\mathit{d}\mathit{x}\mathbf{=}\frac{{\mathbf{e}}^{\mathbf{ax}}\mathbf{}\left(a\sin \mathrm{bx}-b\mathrm{cod}\mathrm{bx}\right)}{{\mathbf{a}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}}$

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

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

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

$$\begin{gathered} \int {\mathrm{e}}^{\left(\mathrm{a}+\mathrm{ib}\right)\mathrm{x}}dx=\frac{{\mathrm{e}}^{\left(\mathrm{a}+\mathrm{ib}\right)\mathrm{x}}}{a+ib} \\ \int {\mathrm{e}}^{\left(\mathrm{a}+\mathrm{ib}\right)\mathrm{x}}dx=\frac{{\mathrm{e}}^{\mathrm{ax}}.{\mathrm{e}}^{\mathrm{ibx}}}{a+ib} \\ \int {\mathrm{e}}^{\left(\mathrm{a}+\mathrm{ib}\right)\mathrm{x}}dx=\frac{{\mathrm{e}}^{ax}\left[\cos \mathrm{bx}+\mathrm{i}\sin \mathrm{bx}\right]}{a+ib} \end{gathered}$$

Multiply the numerator and the denominator with the complex conjugate of

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

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

Using equation (1) and (2).

$\int e^{ax}\cos bxdx+i\int e^{ax}\sin bxdx=\frac{e^{ax}\left(a\cos bx+b\sin bx\right)+i{e}^{ax}\left(a\sin bx-b\cos bx\right)}{a^2+b^2}$

Equate the imaginary part on both sides.

role="math" localid="1658729614203" $\int e^{ax}\sin bxdx=\frac{e^{ax}\left(a\sin bx-bcosbx\right)}{a^2+b^2}$

Therefore, the evaluation of $\int e^{ax}\sin bxdx=\frac{e^{ax}\left(a\sin bx+bcosbx\right)+i{e}^{ax}\left(a\sin bx-b\cos bx\right)}{a^2+b^2}$and the function $\int e^{ax}\sin bxdx=\frac{e^{ax}\left(a\sin bx-bcosbx\right)}{a^2+b^2}$has been showed.