Integral von $$$x \cos{\left(9 x \right)}$$$

Bestimme $$$\int x \cos{\left(9 x \right)}\, dx$$$.

Für das Integral $$$\int{x \cos{\left(9 x \right)} d x}$$$ verwenden Sie die partielle Integration $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Seien $$$\operatorname{u}=x$$$ und $$$\operatorname{dv}=\cos{\left(9 x \right)} dx$$$.

Dann gilt $$$\operatorname{du}=\left(x\right)^{\prime }dx=1 dx$$$ (Rechenschritte siehe ») und $$$\operatorname{v}=\int{\cos{\left(9 x \right)} d x}=\frac{\sin{\left(9 x \right)}}{9}$$$ (Rechenschritte siehe »).

Daher,

$${\color{red}{\int{x \cos{\left(9 x \right)} d x}}}={\color{red}{\left(x \cdot \frac{\sin{\left(9 x \right)}}{9}-\int{\frac{\sin{\left(9 x \right)}}{9} \cdot 1 d x}\right)}}={\color{red}{\left(\frac{x \sin{\left(9 x \right)}}{9} - \int{\frac{\sin{\left(9 x \right)}}{9} d x}\right)}}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=\frac{1}{9}$$$ und $$$f{\left(x \right)} = \sin{\left(9 x \right)}$$$ an:

$$\frac{x \sin{\left(9 x \right)}}{9} - {\color{red}{\int{\frac{\sin{\left(9 x \right)}}{9} d x}}} = \frac{x \sin{\left(9 x \right)}}{9} - {\color{red}{\left(\frac{\int{\sin{\left(9 x \right)} d x}}{9}\right)}}$$

Sei $$$u=9 x$$$.

Dann $$$du=\left(9 x\right)^{\prime }dx = 9 dx$$$ (die Schritte sind » zu sehen), und es gilt $$$dx = \frac{du}{9}$$$.

Das Integral lässt sich umschreiben als

$$\frac{x \sin{\left(9 x \right)}}{9} - \frac{{\color{red}{\int{\sin{\left(9 x \right)} d x}}}}{9} = \frac{x \sin{\left(9 x \right)}}{9} - \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{9} d u}}}}{9}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=\frac{1}{9}$$$ und $$$f{\left(u \right)} = \sin{\left(u \right)}$$$ an:

$$\frac{x \sin{\left(9 x \right)}}{9} - \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{9} d u}}}}{9} = \frac{x \sin{\left(9 x \right)}}{9} - \frac{{\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{9}\right)}}}{9}$$

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

$$\frac{x \sin{\left(9 x \right)}}{9} - \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{81} = \frac{x \sin{\left(9 x \right)}}{9} - \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{81}$$

Zur Erinnerung: $$$u=9 x$$$:

$$\frac{x \sin{\left(9 x \right)}}{9} + \frac{\cos{\left({\color{red}{u}} \right)}}{81} = \frac{x \sin{\left(9 x \right)}}{9} + \frac{\cos{\left({\color{red}{\left(9 x\right)}} \right)}}{81}$$

Daher,

$$\int{x \cos{\left(9 x \right)} d x} = \frac{x \sin{\left(9 x \right)}}{9} + \frac{\cos{\left(9 x \right)}}{81}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{x \cos{\left(9 x \right)} d x} = \frac{x \sin{\left(9 x \right)}}{9} + \frac{\cos{\left(9 x \right)}}{81}+C$$

$$$\int x \cos{\left(9 x \right)}\, dx = \left(\frac{x \sin{\left(9 x \right)}}{9} + \frac{\cos{\left(9 x \right)}}{81}\right) + C$$$A