Integral von $$$2 \sin^{2}{\left(x \right)} - 1$$$

Bestimme $$$\int \left(2 \sin^{2}{\left(x \right)} - 1\right)\, dx$$$.

Gliedweise integrieren:

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

Wenden Sie die Konstantenregel $$$\int c\, dx = c x$$$ mit $$$c=1$$$ an:

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

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

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

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

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

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

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

Gliedweise integrieren:

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

Wenden Sie die Konstantenregel $$$\int c\, dx = c x$$$ mit $$$c=1$$$ an:

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

Sei $$$u=2 x$$$.

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

Somit,

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

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

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

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

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

Zur Erinnerung: $$$u=2 x$$$:

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

Daher,

$$\int{\left(2 \sin^{2}{\left(x \right)} - 1\right)d x} = - \frac{\sin{\left(2 x \right)}}{2}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\left(2 \sin^{2}{\left(x \right)} - 1\right)d x} = - \frac{\sin{\left(2 x \right)}}{2}+C$$

$$$\int \left(2 \sin^{2}{\left(x \right)} - 1\right)\, dx = - \frac{\sin{\left(2 x \right)}}{2} + C$$$A