Integral von $$$\sin{\left(\theta \right)} \cos{\left(\theta \right)}$$$

Bestimme $$$\int \sin{\left(\theta \right)} \cos{\left(\theta \right)}\, d\theta$$$.

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

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

Das Integral wird zu

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

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

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

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

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

Daher,

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

Fügen Sie die Integrationskonstante hinzu:

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

$$$\int \sin{\left(\theta \right)} \cos{\left(\theta \right)}\, d\theta = \frac{\sin^{2}{\left(\theta \right)}}{2} + C$$$A