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

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

Sei $$$u=\cos{\left(x \right)}$$$.

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

Das Integral wird zu

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

Partialbruchzerlegung durchführen (die Schritte sind » zu sehen):

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

Gliedweise integrieren:

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

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

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

Sei $$$v=u - 1$$$.

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

Somit,

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

Das Integral von $$$\frac{1}{v}$$$ ist $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:

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

Zur Erinnerung: $$$v=u - 1$$$:

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

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

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

Sei $$$v=u + 1$$$.

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

Somit,

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

Das Integral von $$$\frac{1}{v}$$$ ist $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:

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

Zur Erinnerung: $$$v=u + 1$$$:

$$\frac{\ln{\left(\left|{u - 1}\right| \right)}}{2} - \frac{\ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{2} = \frac{\ln{\left(\left|{u - 1}\right| \right)}}{2} - \frac{\ln{\left(\left|{{\color{red}{\left(u + 1\right)}}}\right| \right)}}{2}$$

Zur Erinnerung: $$$u=\cos{\left(x \right)}$$$:

$$\frac{\ln{\left(\left|{-1 + {\color{red}{u}}}\right| \right)}}{2} - \frac{\ln{\left(\left|{1 + {\color{red}{u}}}\right| \right)}}{2} = \frac{\ln{\left(\left|{-1 + {\color{red}{\cos{\left(x \right)}}}}\right| \right)}}{2} - \frac{\ln{\left(\left|{1 + {\color{red}{\cos{\left(x \right)}}}}\right| \right)}}{2}$$

Daher,

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

Fügen Sie die Integrationskonstante hinzu:

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

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