Integraal van $$$10 e^{i k n t t_{1}}$$$ met betrekking tot $$$t$$$

Bepaal $$$\int 10 e^{i k n t t_{1}}\, dt$$$.

Pas de constante-veelvoudregel $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ toe met $$$c=10$$$ en $$$f{\left(t \right)} = e^{i k n t t_{1}}$$$:

$${\color{red}{\int{10 e^{i k n t t_{1}} d t}}} = {\color{red}{\left(10 \int{e^{i k n t t_{1}} d t}\right)}}$$

Zij $$$u=i k n t t_{1}$$$.

Dan $$$du=\left(i k n t t_{1}\right)^{\prime }dt = i k n t_{1} dt$$$ (de stappen zijn te zien »), en dan geldt dat $$$dt = - \frac{i du}{k n t_{1}}$$$.

Dus,

$$10 {\color{red}{\int{e^{i k n t t_{1}} d t}}} = 10 {\color{red}{\int{\left(- \frac{i e^{u}}{k n t_{1}}\right)d u}}}$$

Pas de constante-veelvoudregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ toe met $$$c=- \frac{i}{k n t_{1}}$$$ en $$$f{\left(u \right)} = e^{u}$$$:

$$10 {\color{red}{\int{\left(- \frac{i e^{u}}{k n t_{1}}\right)d u}}} = 10 {\color{red}{\left(- \frac{i \int{e^{u} d u}}{k n t_{1}}\right)}}$$

De integraal van de exponentiële functie is $$$\int{e^{u} d u} = e^{u}$$$:

$$- \frac{10 i {\color{red}{\int{e^{u} d u}}}}{k n t_{1}} = - \frac{10 i {\color{red}{e^{u}}}}{k n t_{1}}$$

We herinneren eraan dat $$$u=i k n t t_{1}$$$:

$$- \frac{10 i e^{{\color{red}{u}}}}{k n t_{1}} = - \frac{10 i e^{{\color{red}{i k n t t_{1}}}}}{k n t_{1}}$$

Dus,

$$\int{10 e^{i k n t t_{1}} d t} = - \frac{10 i e^{i k n t t_{1}}}{k n t_{1}}$$

Vereenvoudig:

$$\int{10 e^{i k n t t_{1}} d t} = \frac{10 \left(\sin{\left(k n t t_{1} \right)} - i \cos{\left(k n t t_{1} \right)}\right)}{k n t_{1}}$$

Voeg de integratieconstante toe:

$$\int{10 e^{i k n t t_{1}} d t} = \frac{10 \left(\sin{\left(k n t t_{1} \right)} - i \cos{\left(k n t t_{1} \right)}\right)}{k n t_{1}}+C$$

$$$\int 10 e^{i k n t t_{1}}\, dt = \frac{10 \left(\sin{\left(k n t t_{1} \right)} - i \cos{\left(k n t t_{1} \right)}\right)}{k n t_{1}} + C$$$A