Integral von $$$\frac{1}{\left(1 - \tan{\left(x \right)}\right)^{2} \cos^{2}{\left(x \right)}}$$$

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

Sei $$$u=1 - \tan{\left(x \right)}$$$.

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

Das Integral lässt sich umschreiben als

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

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

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

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

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

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

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

Daher,

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

Vereinfachen:

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

Fügen Sie die Integrationskonstante hinzu:

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

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