Answer
a) $\color{blue}{\displaystyle \int e^{(1+i)x}\ dx = \frac{1}{2}e^x(\cos x +\sin x) + i\frac{1}{2}e^x(\sin x -\cos x) + C_z },$
where $C_z = C_R + iC_I$ is a (complex number) constant of integration.
b)
i) Comparing real parts: $\color{blue}{\displaystyle \int e^x\cos x\ dx = \frac{1}{2}e^x(\cos x + \sin x) + C_R}$
ii) Comparing imaginary parts: $\color{blue}{\displaystyle \int e^x\sin x\ dx = \frac{1}{2}e^x(\sin - \cos x) + C_I}$
where $C_R$ and $C_I$ are (real-numbers) constants of integration.
c) Result b.ii) is equivalent to the result obtained in Example 7.1.4 where the technique of integration by parts was used to evaluate $\displaystyle \int e^x\sin x\ dx$.
Work Step by Step
a)
$\begin{align*}
\int e^{(1+i)x}\ dx &= \frac{e^{(1+i)x}}{1+i} + C_z \quad \left[\ \text{since}\ \int e^{ax}\ dx = \frac{e^{ax}}{a} + C\ \right]\\
& \qquad\qquad \text{where}\ C_z\ \text{is an arbitrary complex number constant of intgeration} \\
&= \frac{e^{(1+i)x}}{1+i} \cdot \frac{1-i}{1-i} + C_z \\
&= \frac{e^{(1+i)x}(1-i)}{1^2-i^2} + C_z \\
&= \frac{e^{x+ix}(1-i)}{1-(-1)} + C_z \\
&= \frac{e^xe^{ix}(1-i)}{2} +C_z \\
&= \frac{1}{2}e^x(\cos x + i\sin x)(1-i) + C_z \quad [\text{Euler's Formula}: e^{ix} = \cos x + i\sin x] \\
&= \frac{1}{2}e^x\left((\cos x)(1-i) + (i\sin x)(1-i)\right) + C_z \\
&= \frac{1}{2}e^x\left(\cos x - i\cos x + i\sin x - i^2\sin x\right) + C_z \\
&= \frac{1}{2}e^x\left(\cos x - i\cos x + i\sin x - (-1)\sin x\right) + C_z \\
&= \frac{1}{2}e^x\left((\cos x +\sin x) + i(-\cos x + \sin x)\right) + C_z \\
\int e^{(1+i)x}\ dx &= \frac{1}{2}e^x(\cos x +\sin x) + i\frac{1}{2}e^x(\sin x -\cos x) + C_z \qquad \text{(Eq. 1)}
\end{align*}
$
b) Note that by using Euler's Formula for $e^{ix}$, we can rewrite the integral as
$\begin{align*}
\int e^{(1+i)x}\ dx &= \int e^{x+ix}\ dx \\
&= \int e^xe^{ix}\ dx \\
&= \int e^x(\cos x + i\sin x)\ dx \\
\int e^{(1+i)x}\ dx &= \int e^x\cos x\ dx + i\int e^x\sin x\ dx \qquad \text{(Eq. 2)}
\end{align*}$
Thus, from Eq.2,
* Real part of $\displaystyle \int e^{(1+i)x}\ dx\ \text{is}\ \int e^x\cos x\ dx \qquad \text{(Eq. 3a)}$
* Imaginary part of $\displaystyle \int e^{(1+i)x}\ dx\ \text{is}\ \int e^x\sin x\ dx \qquad \text{(Eq. 3b)}$
Returning to part a), we can write Eq. 1 as
$\begin{align*}
\int e^{(1+i)x}\ dx &= \frac{1}{2}e^x(\cos x +\sin x) + i\left(\frac{1}{2}e^x(\sin x -\cos x)\right) + C_R + iC_I \\
& \qquad\qquad \text{where}\ C_R,\ \text{and}\ C_I\ \text{are, respectively, the real and imaginary parts}\ C_z \\
\int e^{(1+i)x}\ dx &= \frac{1}{2}e^x(\cos x +\sin x) + C_R + i\left(\frac{1}{2}e^x(\sin x - \cos x) + C_I\right)
\end{align*}$
Thus, from a), we also have,
* Real part of $\displaystyle \int e^{(1+i)x}\ dx\ \text{is}\ \frac{1}{2}e^x(\cos x + \sin x) + C_R \qquad \text{(Eq. 4a)}$
* Imaginary part of $\displaystyle \int e^{(1+i)x}\ dx\ \text{is}\ \frac{1}{2}e^x(\sin - \cos x) + C_I \qquad \text{(Eq. 4b)}$
Comparing Eq. 3a with Eq. 4a (comparing the real part of the indefinite integral and that of the antiderivative), we have
$\displaystyle\color{blue}{\int e^x\cos x\ dx = \frac{1}{2}e^x(\cos x +\sin x) + C_R}$.
Similarly, comparing Eq. 3b with Eq. 4b (comparing the imaginary parts of the indefinite integral and that of the antiderivative), we have
$\displaystyle\color{blue}{\int e^x\sin x\ dx = \frac{1}{2}e^x(\sin x - \cos x) + C_I}$.
c) The result obtained by comparing the imaginary parts of the indefinite integral and its antiderivative is equivalent to that obtained in Example 7.1.4 where the technique of integration by parts was used to evaluate $\displaystyle \int e^x\sin x\ dx$.
