Integral von $$$\sin{\left(\sin{\left(x \right)} \right)} \cos{\left(x \right)}$$$

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

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

Dann $$$du=\left(\sin{\left(x \right)}\right)^{\prime }dx = \cos{\left(x \right)} dx$$$ (die Schritte sind » zu sehen), und es gilt $$$\cos{\left(x \right)} dx = du$$$.

Somit,

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

Das Integral des Sinus lautet $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$:

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

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

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

Daher,

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

Fügen Sie die Integrationskonstante hinzu:

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

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