Integraal van $$$\sin{\left(\sin{\left(x \right)} \right)} \cos{\left(x \right)}$$$

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

Zij $$$u=\sin{\left(x \right)}$$$.

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

De integraal kan worden herschreven als

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

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 $$$u=\sin{\left(x \right)}$$$:

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

Dus,

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

Voeg de integratieconstante toe:

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

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