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

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

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

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

Daher,

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

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

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

Schreiben Sie den Integranden in Bezug auf die Kosekans um.:

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

Das Integral von $$$\csc^{2}{\left(u \right)}$$$ ist $$$\int{\csc^{2}{\left(u \right)} d u} = - \cot{\left(u \right)}$$$:

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

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

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

Daher,

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

Fügen Sie die Integrationskonstante hinzu:

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

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