Integral de $$$10 e^{i k n t t_{1}}$$$ em relação a $$$t$$$

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

Aplique a regra do múltiplo constante $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ usando $$$c=10$$$ e $$$f{\left(t \right)} = e^{i k n t t_{1}}$$$:

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

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

Então $$$du=\left(i k n t t_{1}\right)^{\prime }dt = i k n t_{1} dt$$$ (veja os passos »), e obtemos $$$dt = - \frac{i du}{k n t_{1}}$$$.

A integral pode ser reescrita como

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

Aplique a regra do múltiplo constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ usando $$$c=- \frac{i}{k n t_{1}}$$$ e $$$f{\left(u \right)} = e^{u}$$$:

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

A integral da função exponencial é $$$\int{e^{u} d u} = e^{u}$$$:

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

Recorde que $$$u=i k n t t_{1}$$$:

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

Portanto,

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

Simplifique:

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

Adicione a constante de integração:

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

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