Integral von $$$\cot{\left(t \right)}$$$

Bestimme $$$\int \cot{\left(t \right)}\, dt$$$.

Schreibe den Kotangens als $$$\cot\left(t\right)=\frac{\cos\left(t\right)}{\sin\left(t\right)}$$$ um:

$${\color{red}{\int{\cot{\left(t \right)} d t}}} = {\color{red}{\int{\frac{\cos{\left(t \right)}}{\sin{\left(t \right)}} d t}}}$$

Sei $$$u=\sin{\left(t \right)}$$$.

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

Also,

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

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

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

Zur Erinnerung: $$$u=\sin{\left(t \right)}$$$:

$$\ln{\left(\left|{{\color{red}{u}}}\right| \right)} = \ln{\left(\left|{{\color{red}{\sin{\left(t \right)}}}}\right| \right)}$$

Daher,

$$\int{\cot{\left(t \right)} d t} = \ln{\left(\left|{\sin{\left(t \right)}}\right| \right)}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\cot{\left(t \right)} d t} = \ln{\left(\left|{\sin{\left(t \right)}}\right| \right)}+C$$

$$$\int \cot{\left(t \right)}\, dt = \ln\left(\left|{\sin{\left(t \right)}}\right|\right) + C$$$A