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

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

Sei $$$u=9^{x}$$$.

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

Daher,

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

Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=\frac{1}{2 \ln{\left(3 \right)}}$$$ und $$$f{\left(u \right)} = \sin{\left(u \right)}$$$ an:

$${\color{red}{\int{\frac{\sin{\left(u \right)}}{2 \ln{\left(3 \right)}} d u}}} = {\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{2 \ln{\left(3 \right)}}\right)}}$$

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

$$\frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{2 \ln{\left(3 \right)}} = \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{2 \ln{\left(3 \right)}}$$

Zur Erinnerung: $$$u=9^{x}$$$:

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

Daher,

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

Fügen Sie die Integrationskonstante hinzu:

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

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