Integral von $$$x^{3} \sin{\left(x \right)}$$$

Bestimme $$$\int x^{3} \sin{\left(x \right)}\, dx$$$.

Für das Integral $$$\int{x^{3} \sin{\left(x \right)} d x}$$$ verwenden Sie die partielle Integration $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Seien $$$\operatorname{u}=x^{3}$$$ und $$$\operatorname{dv}=\sin{\left(x \right)} dx$$$.

Dann gilt $$$\operatorname{du}=\left(x^{3}\right)^{\prime }dx=3 x^{2} dx$$$ (Rechenschritte siehe ») und $$$\operatorname{v}=\int{\sin{\left(x \right)} d x}=- \cos{\left(x \right)}$$$ (Rechenschritte siehe »).

Das Integral wird zu

$${\color{red}{\int{x^{3} \sin{\left(x \right)} d x}}}={\color{red}{\left(x^{3} \cdot \left(- \cos{\left(x \right)}\right)-\int{\left(- \cos{\left(x \right)}\right) \cdot 3 x^{2} d x}\right)}}={\color{red}{\left(- x^{3} \cos{\left(x \right)} - \int{\left(- 3 x^{2} \cos{\left(x \right)}\right)d x}\right)}}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=-3$$$ und $$$f{\left(x \right)} = x^{2} \cos{\left(x \right)}$$$ an:

$$- x^{3} \cos{\left(x \right)} - {\color{red}{\int{\left(- 3 x^{2} \cos{\left(x \right)}\right)d x}}} = - x^{3} \cos{\left(x \right)} - {\color{red}{\left(- 3 \int{x^{2} \cos{\left(x \right)} d x}\right)}}$$

Für das Integral $$$\int{x^{2} \cos{\left(x \right)} d x}$$$ verwenden Sie die partielle Integration $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Seien $$$\operatorname{u}=x^{2}$$$ und $$$\operatorname{dv}=\cos{\left(x \right)} dx$$$.

Dann gilt $$$\operatorname{du}=\left(x^{2}\right)^{\prime }dx=2 x dx$$$ (Rechenschritte siehe ») und $$$\operatorname{v}=\int{\cos{\left(x \right)} d x}=\sin{\left(x \right)}$$$ (Rechenschritte siehe »).

Das Integral lässt sich umschreiben als

$$- x^{3} \cos{\left(x \right)} + 3 {\color{red}{\int{x^{2} \cos{\left(x \right)} d x}}}=- x^{3} \cos{\left(x \right)} + 3 {\color{red}{\left(x^{2} \cdot \sin{\left(x \right)}-\int{\sin{\left(x \right)} \cdot 2 x d x}\right)}}=- x^{3} \cos{\left(x \right)} + 3 {\color{red}{\left(x^{2} \sin{\left(x \right)} - \int{2 x \sin{\left(x \right)} d x}\right)}}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=2$$$ und $$$f{\left(x \right)} = x \sin{\left(x \right)}$$$ an:

$$- x^{3} \cos{\left(x \right)} + 3 x^{2} \sin{\left(x \right)} - 3 {\color{red}{\int{2 x \sin{\left(x \right)} d x}}} = - x^{3} \cos{\left(x \right)} + 3 x^{2} \sin{\left(x \right)} - 3 {\color{red}{\left(2 \int{x \sin{\left(x \right)} d x}\right)}}$$

Für das Integral $$$\int{x \sin{\left(x \right)} d x}$$$ verwenden Sie die partielle Integration $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Seien $$$\operatorname{u}=x$$$ und $$$\operatorname{dv}=\sin{\left(x \right)} dx$$$.

Dann gilt $$$\operatorname{du}=\left(x\right)^{\prime }dx=1 dx$$$ (Rechenschritte siehe ») und $$$\operatorname{v}=\int{\sin{\left(x \right)} d x}=- \cos{\left(x \right)}$$$ (Rechenschritte siehe »).

Somit,

$$- x^{3} \cos{\left(x \right)} + 3 x^{2} \sin{\left(x \right)} - 6 {\color{red}{\int{x \sin{\left(x \right)} d x}}}=- x^{3} \cos{\left(x \right)} + 3 x^{2} \sin{\left(x \right)} - 6 {\color{red}{\left(x \cdot \left(- \cos{\left(x \right)}\right)-\int{\left(- \cos{\left(x \right)}\right) \cdot 1 d x}\right)}}=- x^{3} \cos{\left(x \right)} + 3 x^{2} \sin{\left(x \right)} - 6 {\color{red}{\left(- x \cos{\left(x \right)} - \int{\left(- \cos{\left(x \right)}\right)d x}\right)}}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=-1$$$ und $$$f{\left(x \right)} = \cos{\left(x \right)}$$$ an:

$$- x^{3} \cos{\left(x \right)} + 3 x^{2} \sin{\left(x \right)} + 6 x \cos{\left(x \right)} + 6 {\color{red}{\int{\left(- \cos{\left(x \right)}\right)d x}}} = - x^{3} \cos{\left(x \right)} + 3 x^{2} \sin{\left(x \right)} + 6 x \cos{\left(x \right)} + 6 {\color{red}{\left(- \int{\cos{\left(x \right)} d x}\right)}}$$

Das Integral des Kosinus ist $$$\int{\cos{\left(x \right)} d x} = \sin{\left(x \right)}$$$:

$$- x^{3} \cos{\left(x \right)} + 3 x^{2} \sin{\left(x \right)} + 6 x \cos{\left(x \right)} - 6 {\color{red}{\int{\cos{\left(x \right)} d x}}} = - x^{3} \cos{\left(x \right)} + 3 x^{2} \sin{\left(x \right)} + 6 x \cos{\left(x \right)} - 6 {\color{red}{\sin{\left(x \right)}}}$$

Daher,

$$\int{x^{3} \sin{\left(x \right)} d x} = - x^{3} \cos{\left(x \right)} + 3 x^{2} \sin{\left(x \right)} + 6 x \cos{\left(x \right)} - 6 \sin{\left(x \right)}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{x^{3} \sin{\left(x \right)} d x} = - x^{3} \cos{\left(x \right)} + 3 x^{2} \sin{\left(x \right)} + 6 x \cos{\left(x \right)} - 6 \sin{\left(x \right)}+C$$

$$$\int x^{3} \sin{\left(x \right)}\, dx = \left(- x^{3} \cos{\left(x \right)} + 3 x^{2} \sin{\left(x \right)} + 6 x \cos{\left(x \right)} - 6 \sin{\left(x \right)}\right) + C$$$A