Chapter 9.3, Problem 50E

a.

To determine

To prove: To prove the confirmation of the antiderivative formula.

Explanation of Solution

Given:

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

Calculation:

Given ${\displaystyle \int e^{(a+ib)x}dx}=\frac{a-ib}{a^2+b^2}{e}^{(a+ib)x}+C$

Differentiating on both the side with respect to $x$ .

  $\frac{d}{dx}\left[{\displaystyle \int e^{(a+ib)x}dx}\right]=\frac{d}{dx}\left[\frac{a-ib}{a^2+b^2}{e}^{(a+ib)x}+C\right]$

Taking left hand side

  $\frac{d}{dx}\left[{\displaystyle \int e^{(a+ib)x}dx}\right]=e^{(a+ib)x}$

Taking right hand side

  $\begin{array}{c}\frac{d}{dx}\left[\frac{a-ib}{a^2+b^2}{e}^{(a+ib)x}+C\right]=\frac{a-ib}{a^2+b^2}{e}^{(a+ib)x}\\ =\frac{a^2-{\left(ib\right)}^2}{a^2+b^2}{e}^{(a+ib)x}\\ =\frac{a^2+b^2}{a^2+b^2}{e}^{(a+ib)x}\\ =e^{\left(a+bi\right)x}\end{array}$

This proved that it is a confirm antiderivative formula.

b.

To determine

To evaluate: To evaluate ${\displaystyle \int e^{ax}\cos bxdx}$ and ${\displaystyle \int e^{ax}\sin bxdx}$ using the fact that two complex numbers $a+ib$ and $c+id$ where $a=b$ and $c=d$ .

Explanation of Solution

Given:

The given function $e^{ax}(\cos bx+i\sin bx)=e^{(a+ib)x}$

Calculation:

Integrating both the sides with respect to x.

  $\begin{array}{l}{\displaystyle \int e^{ax}(\cos bx+i\sin bx)}dx={\displaystyle \int e^{(a+ib)x}}dx\\ {\displaystyle \int e^{ax}(\cos bx+i\sin bx)}dx=\frac{a-ib}{a^2+b^2}{e}^{(a+ib)x}\\ {\displaystyle \int e^{ax}(\cos bx+i\sin bx)}dx=\frac{a-ib}{a^2+b^2}{e}^{ax}(\cos bx+i\sin bx)\\ {\displaystyle \int e^{ax}(\cos bx+i\sin bx)}dx=\frac{1}{a^2+b^2}{e}^{ax}(a\cos bx+b\sin bx-bi\cos x+ai\sin bx)\\ {\displaystyle \int e^{ax}(\cos bx+i\sin bx)}dx=\frac{1}{a^2+b^2}{e}^{ax}((a\cos bx+b\sin bx)+i(-a\cos x+a\sin bx)\end{array}$

Separating the real and imaginary parts,

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

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