Integral von $$$\tan{\left(x \right)} + 1 - \frac{1}{\tan{\left(x \right)}}$$$

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

Gliedweise integrieren:

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

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

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

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

Dann gelten $$$x=\operatorname{atan}{\left(u \right)}$$$ und $$$dx=\left(\operatorname{atan}{\left(u \right)}\right)^{\prime }du = \frac{du}{u^{2} + 1}$$$ (die Schritte sind » zu sehen).

Daher,

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

Sei $$$v=u^{2} + 1$$$.

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

Das Integral wird zu

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

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

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

Partialbruchzerlegung durchführen (die Schritte sind » zu sehen):

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

Gliedweise integrieren:

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

Sei $$$w=v - 1$$$.

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

Das Integral lässt sich umschreiben als

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

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

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

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

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

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

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

Zur Erinnerung: $$$v=u^{2} + 1$$$:

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

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

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

Schreibe die Tangente als $$$\tan\left(x\right)=\frac{\sin\left(x\right)}{\cos\left(x\right)}$$$ um:

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

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 lässt sich umschreiben als

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

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

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

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

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

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

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

Daher,

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

Vereinfachen:

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

Fügen Sie die Integrationskonstante hinzu:

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

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