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

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

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

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

Somit,

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

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

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

Schreibe $$$1$$$ als $$$\sin^2\left(\frac{ u }{2}\right)+\cos^2\left(\frac{ u }{2}\right)$$$ um und wende die Doppelwinkelformel für den Sinus $$$\sin\left( u \right)=2\sin\left(\frac{ u }{2}\right)\cos\left(\frac{ u }{2}\right)$$$ an.:

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

Quadratische Ergänzung (Schritte können » eingesehen werden):

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

Multipliziere Zähler und Nenner mit $$$\sec^2\left(\frac{ u }{2}\right)$$$:

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

Sei $$$v=\tan{\left(\frac{u}{2} \right)} - 1$$$.

Dann $$$dv=\left(\tan{\left(\frac{u}{2} \right)} - 1\right)^{\prime }du = \frac{\sec^{2}{\left(\frac{u}{2} \right)}}{2} du$$$ (die Schritte sind » zu sehen), und es gilt $$$\sec^{2}{\left(\frac{u}{2} \right)} du = 2 dv$$$.

Das Integral lässt sich umschreiben als

$$- \frac{{\color{red}{\int{\left(- \frac{\sec^{2}{\left(\frac{u}{2} \right)}}{\left(\tan{\left(\frac{u}{2} \right)} - 1\right)^{2}}\right)d u}}}}{2} = - \frac{{\color{red}{\int{\left(- \frac{2}{v^{2}}\right)d v}}}}{2}$$

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

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

Wenden Sie die Potenzregel $$$\int v^{n}\, dv = \frac{v^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=-2$$$ an:

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

Zur Erinnerung: $$$v=\tan{\left(\frac{u}{2} \right)} - 1$$$:

$$- {\color{red}{v}}^{-1} = - {\color{red}{\left(\tan{\left(\frac{u}{2} \right)} - 1\right)}}^{-1}$$

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

$$- \left(-1 + \tan{\left(\frac{{\color{red}{u}}}{2} \right)}\right)^{-1} = - \left(-1 + \tan{\left(\frac{{\color{red}{\left(2 x\right)}}}{2} \right)}\right)^{-1}$$

Daher,

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

Fügen Sie die Integrationskonstante hinzu:

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

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