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

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

Sei $$$u=\frac{x}{2}$$$.

Dann $$$du=\left(\frac{x}{2}\right)^{\prime }dx = \frac{dx}{2}$$$ (die Schritte sind » zu sehen), und es gilt $$$dx = 2 du$$$.

Somit,

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

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

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

Wende die Potenzreduktionsformel $$$\sin^{2}{\left(\alpha \right)} = \frac{1}{2} - \frac{\cos{\left(2 \alpha \right)}}{2}$$$ mit $$$\alpha= u $$$ an:

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

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

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

Gliedweise integrieren:

$${\color{red}{\int{\left(1 - \cos{\left(2 u \right)}\right)d u}}} = {\color{red}{\left(\int{1 d u} - \int{\cos{\left(2 u \right)} d u}\right)}}$$

Wenden Sie die Konstantenregel $$$\int c\, du = c u$$$ mit $$$c=1$$$ an:

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

Sei $$$v=2 u$$$.

Dann $$$dv=\left(2 u\right)^{\prime }du = 2 du$$$ (die Schritte sind » zu sehen), und es gilt $$$du = \frac{dv}{2}$$$.

Daher,

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

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

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

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

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

Zur Erinnerung: $$$v=2 u$$$:

$$u - \frac{\sin{\left({\color{red}{v}} \right)}}{2} = u - \frac{\sin{\left({\color{red}{\left(2 u\right)}} \right)}}{2}$$

Zur Erinnerung: $$$u=\frac{x}{2}$$$:

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

Daher,

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

Vereinfachen:

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

Fügen Sie die Integrationskonstante hinzu:

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

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