Integral von $$$\frac{\sin{\left(x \right)} \cos{\left(x \right)}}{2}$$$

Bestimme $$$\int \frac{\sin{\left(x \right)} \cos{\left(x \right)}}{2}\, dx$$$.

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

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

Sei $$$u=\sin{\left(x \right)}$$$.

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

Somit,

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

Wenden Sie die Potenzregel $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=1$$$ an:

$$\frac{{\color{red}{\int{u d u}}}}{2}=\frac{{\color{red}{\frac{u^{1 + 1}}{1 + 1}}}}{2}=\frac{{\color{red}{\left(\frac{u^{2}}{2}\right)}}}{2}$$

Zur Erinnerung: $$$u=\sin{\left(x \right)}$$$:

$$\frac{{\color{red}{u}}^{2}}{4} = \frac{{\color{red}{\sin{\left(x \right)}}}^{2}}{4}$$

Daher,

$$\int{\frac{\sin{\left(x \right)} \cos{\left(x \right)}}{2} d x} = \frac{\sin^{2}{\left(x \right)}}{4}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{\sin{\left(x \right)} \cos{\left(x \right)}}{2} d x} = \frac{\sin^{2}{\left(x \right)}}{4}+C$$

$$$\int \frac{\sin{\left(x \right)} \cos{\left(x \right)}}{2}\, dx = \frac{\sin^{2}{\left(x \right)}}{4} + C$$$A