Integral de $$$\frac{j_{0} \sin{\left(k^{2} t \right)}}{k}$$$ em relação a $$$t$$$

Encontre $$$\int \frac{j_{0} \sin{\left(k^{2} t \right)}}{k}\, dt$$$.

Aplique a regra do múltiplo constante $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ usando $$$c=\frac{j_{0}}{k}$$$ e $$$f{\left(t \right)} = \sin{\left(k^{2} t \right)}$$$:

$${\color{red}{\int{\frac{j_{0} \sin{\left(k^{2} t \right)}}{k} d t}}} = {\color{red}{\frac{j_{0} \int{\sin{\left(k^{2} t \right)} d t}}{k}}}$$

Seja $$$u=k^{2} t$$$.

Então $$$du=\left(k^{2} t\right)^{\prime }dt = k^{2} dt$$$ (veja os passos »), e obtemos $$$dt = \frac{du}{k^{2}}$$$.

Assim,

$$\frac{j_{0} {\color{red}{\int{\sin{\left(k^{2} t \right)} d t}}}}{k} = \frac{j_{0} {\color{red}{\int{\frac{\sin{\left(u \right)}}{k^{2}} d u}}}}{k}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ usando $$$c=\frac{1}{k^{2}}$$$ e $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

$$\frac{j_{0} {\color{red}{\int{\frac{\sin{\left(u \right)}}{k^{2}} d u}}}}{k} = \frac{j_{0} {\color{red}{\frac{\int{\sin{\left(u \right)} d u}}{k^{2}}}}}{k}$$

A integral do seno é $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$:

$$\frac{j_{0} {\color{red}{\int{\sin{\left(u \right)} d u}}}}{k^{3}} = \frac{j_{0} {\color{red}{\left(- \cos{\left(u \right)}\right)}}}{k^{3}}$$

Recorde que $$$u=k^{2} t$$$:

$$- \frac{j_{0} \cos{\left({\color{red}{u}} \right)}}{k^{3}} = - \frac{j_{0} \cos{\left({\color{red}{k^{2} t}} \right)}}{k^{3}}$$

Portanto,

$$\int{\frac{j_{0} \sin{\left(k^{2} t \right)}}{k} d t} = - \frac{j_{0} \cos{\left(k^{2} t \right)}}{k^{3}}$$

Adicione a constante de integração:

$$\int{\frac{j_{0} \sin{\left(k^{2} t \right)}}{k} d t} = - \frac{j_{0} \cos{\left(k^{2} t \right)}}{k^{3}}+C$$

$$$\int \frac{j_{0} \sin{\left(k^{2} t \right)}}{k}\, dt = - \frac{j_{0} \cos{\left(k^{2} t \right)}}{k^{3}} + C$$$A