Integraal van $$$\sin{\left(2 t - 2 x \right)}$$$ met betrekking tot $$$x$$$

Bepaal $$$\int \sin{\left(2 t - 2 x \right)}\, dx$$$.

Zij $$$u=2 t - 2 x$$$.

Dan $$$du=\left(2 t - 2 x\right)^{\prime }dx = - 2 dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$dx = - \frac{du}{2}$$$.

Dus,

$${\color{red}{\int{\sin{\left(2 t - 2 x \right)} d x}}} = {\color{red}{\int{\left(- \frac{\sin{\left(u \right)}}{2}\right)d u}}}$$

Pas de constante-veelvoudregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ toe met $$$c=- \frac{1}{2}$$$ en $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

$${\color{red}{\int{\left(- \frac{\sin{\left(u \right)}}{2}\right)d u}}} = {\color{red}{\left(- \frac{\int{\sin{\left(u \right)} d u}}{2}\right)}}$$

De integraal van de sinus is $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$:

$$- \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{2} = - \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{2}$$

We herinneren eraan dat $$$u=2 t - 2 x$$$:

$$\frac{\cos{\left({\color{red}{u}} \right)}}{2} = \frac{\cos{\left({\color{red}{\left(2 t - 2 x\right)}} \right)}}{2}$$

Dus,

$$\int{\sin{\left(2 t - 2 x \right)} d x} = \frac{\cos{\left(2 t - 2 x \right)}}{2}$$

Vereenvoudig:

$$\int{\sin{\left(2 t - 2 x \right)} d x} = \frac{\cos{\left(2 \left(- t + x\right) \right)}}{2}$$

Voeg de integratieconstante toe:

$$\int{\sin{\left(2 t - 2 x \right)} d x} = \frac{\cos{\left(2 \left(- t + x\right) \right)}}{2}+C$$

$$$\int \sin{\left(2 t - 2 x \right)}\, dx = \frac{\cos{\left(2 \left(- t + x\right) \right)}}{2} + C$$$A