Integraal van $$$\sin{\left(x + y \right)}$$$ met betrekking tot $$$x$$$

Bepaal $$$\int \sin{\left(x + y \right)}\, dx$$$.

Zij $$$u=x + y$$$.

Dan $$$du=\left(x + y\right)^{\prime }dx = 1 dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$dx = du$$$.

Dus,

$${\color{red}{\int{\sin{\left(x + y \right)} d x}}} = {\color{red}{\int{\sin{\left(u \right)} d u}}}$$

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

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

We herinneren eraan dat $$$u=x + y$$$:

$$- \cos{\left({\color{red}{u}} \right)} = - \cos{\left({\color{red}{\left(x + y\right)}} \right)}$$

Dus,

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

Voeg de integratieconstante toe:

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

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