Integral von $$$x e^{2} \sin{\left(3 x \right)}$$$

Bestimme $$$\int x e^{2} \sin{\left(3 x \right)}\, dx$$$.

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

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

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

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

Dann gilt $$$\operatorname{du}=\left(x\right)^{\prime }dx=1 dx$$$ (Rechenschritte siehe ») und $$$\operatorname{v}=\int{\sin{\left(3 x \right)} d x}=- \frac{\cos{\left(3 x \right)}}{3}$$$ (Rechenschritte siehe »).

Somit,

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

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=- \frac{1}{3}$$$ und $$$f{\left(x \right)} = \cos{\left(3 x \right)}$$$ an:

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

Sei $$$u=3 x$$$.

Dann $$$du=\left(3 x\right)^{\prime }dx = 3 dx$$$ (die Schritte sind » zu sehen), und es gilt $$$dx = \frac{du}{3}$$$.

Das Integral lässt sich umschreiben als

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

Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=\frac{1}{3}$$$ und $$$f{\left(u \right)} = \cos{\left(u \right)}$$$ an:

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

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

$$e^{2} \left(- \frac{x \cos{\left(3 x \right)}}{3} + \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{9}\right) = e^{2} \left(- \frac{x \cos{\left(3 x \right)}}{3} + \frac{{\color{red}{\sin{\left(u \right)}}}}{9}\right)$$

Zur Erinnerung: $$$u=3 x$$$:

$$e^{2} \left(- \frac{x \cos{\left(3 x \right)}}{3} + \frac{\sin{\left({\color{red}{u}} \right)}}{9}\right) = e^{2} \left(- \frac{x \cos{\left(3 x \right)}}{3} + \frac{\sin{\left({\color{red}{\left(3 x\right)}} \right)}}{9}\right)$$

Daher,

$$\int{x e^{2} \sin{\left(3 x \right)} d x} = \left(- \frac{x \cos{\left(3 x \right)}}{3} + \frac{\sin{\left(3 x \right)}}{9}\right) e^{2}$$

Vereinfachen:

$$\int{x e^{2} \sin{\left(3 x \right)} d x} = \frac{\left(- 3 x \cos{\left(3 x \right)} + \sin{\left(3 x \right)}\right) e^{2}}{9}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{x e^{2} \sin{\left(3 x \right)} d x} = \frac{\left(- 3 x \cos{\left(3 x \right)} + \sin{\left(3 x \right)}\right) e^{2}}{9}+C$$

$$$\int x e^{2} \sin{\left(3 x \right)}\, dx = \frac{\left(- 3 x \cos{\left(3 x \right)} + \sin{\left(3 x \right)}\right) e^{2}}{9} + C$$$A