Integral von $$$- \frac{\cos{\left(4 x \right)}}{4}$$$

Bestimme $$$\int \left(- \frac{\cos{\left(4 x \right)}}{4}\right)\, dx$$$.

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=- \frac{1}{4}$$$ und $$$f{\left(x \right)} = \cos{\left(4 x \right)}$$$ an:

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

Sei $$$u=4 x$$$.

Dann $$$du=\left(4 x\right)^{\prime }dx = 4 dx$$$ (die Schritte sind » zu sehen), und es gilt $$$dx = \frac{du}{4}$$$.

Also,

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

Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=\frac{1}{4}$$$ und $$$f{\left(u \right)} = \cos{\left(u \right)}$$$ an:

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

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

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

Zur Erinnerung: $$$u=4 x$$$:

$$- \frac{\sin{\left({\color{red}{u}} \right)}}{16} = - \frac{\sin{\left({\color{red}{\left(4 x\right)}} \right)}}{16}$$

Daher,

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

Fügen Sie die Integrationskonstante hinzu:

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

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