Integral von $$$\frac{\sin{\left(16 x \right)}}{\cos{\left(8 x \right)}}$$$

Bestimme $$$\int \frac{\sin{\left(16 x \right)}}{\cos{\left(8 x \right)}}\, dx$$$.

Sei $$$u=8 x$$$.

Dann $$$du=\left(8 x\right)^{\prime }dx = 8 dx$$$ (die Schritte sind » zu sehen), und es gilt $$$dx = \frac{du}{8}$$$.

Das Integral lässt sich umschreiben als

$${\color{red}{\int{\frac{\sin{\left(16 x \right)}}{\cos{\left(8 x \right)}} d x}}} = {\color{red}{\int{\frac{\sin{\left(2 u \right)}}{8 \cos{\left(u \right)}} d u}}}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=\frac{1}{8}$$$ und $$$f{\left(u \right)} = \frac{\sin{\left(2 u \right)}}{\cos{\left(u \right)}}$$$ an:

$${\color{red}{\int{\frac{\sin{\left(2 u \right)}}{8 \cos{\left(u \right)}} d u}}} = {\color{red}{\left(\frac{\int{\frac{\sin{\left(2 u \right)}}{\cos{\left(u \right)}} d u}}{8}\right)}}$$

Schreiben Sie den Integranden um:

$$\frac{{\color{red}{\int{\frac{\sin{\left(2 u \right)}}{\cos{\left(u \right)}} d u}}}}{8} = \frac{{\color{red}{\int{2 \sin{\left(u \right)} d u}}}}{8}$$

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

$$\frac{{\color{red}{\int{2 \sin{\left(u \right)} d u}}}}{8} = \frac{{\color{red}{\left(2 \int{\sin{\left(u \right)} d u}\right)}}}{8}$$

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

$$\frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{4} = \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{4}$$

Zur Erinnerung: $$$u=8 x$$$:

$$- \frac{\cos{\left({\color{red}{u}} \right)}}{4} = - \frac{\cos{\left({\color{red}{\left(8 x\right)}} \right)}}{4}$$

Daher,

$$\int{\frac{\sin{\left(16 x \right)}}{\cos{\left(8 x \right)}} d x} = - \frac{\cos{\left(8 x \right)}}{4}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{\sin{\left(16 x \right)}}{\cos{\left(8 x \right)}} d x} = - \frac{\cos{\left(8 x \right)}}{4}+C$$

$$$\int \frac{\sin{\left(16 x \right)}}{\cos{\left(8 x \right)}}\, dx = - \frac{\cos{\left(8 x \right)}}{4} + C$$$A