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

Bestimme $$$\int t^{3} \sin{\left(t \right)}\, dt$$$.

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

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

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

Das Integral lässt sich umschreiben als

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

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

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

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

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

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

Also,

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

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

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

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

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

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

Daher,

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

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

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

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

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

Daher,

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

Fügen Sie die Integrationskonstante hinzu:

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

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