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

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

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 $$$dx = 2 du$$$.

Das Integral wird zu

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

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

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

Gliedweise integrieren:

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

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

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

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

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

Somit,

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

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

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

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

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

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

$$2 u + \cos{\left({\color{red}{v}} \right)} = 2 u + \cos{\left({\color{red}{\left(2 u\right)}} \right)}$$

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

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

Daher,

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

Fügen Sie die Integrationskonstante hinzu:

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

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