Integralen av $$$1 - \cot{\left(x \right)}$$$

Bestäm $$$\int \left(1 - \cot{\left(x \right)}\right)\, dx$$$.

Integrera termvis:

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

Tillämpa konstantregeln $$$\int c\, dx = c x$$$ med $$$c=1$$$:

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

Skriv om kotangensen som $$$\cot\left(x\right)=\frac{\cos\left(x\right)}{\sin\left(x\right)}$$$:

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

Låt $$$u=\sin{\left(x \right)}$$$ vara.

Då $$$du=\left(\sin{\left(x \right)}\right)^{\prime }dx = \cos{\left(x \right)} dx$$$ (stegen kan ses »), och vi har att $$$\cos{\left(x \right)} dx = du$$$.

Integralen blir

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

Integralen av $$$\frac{1}{u}$$$ är $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

Kom ihåg att $$$u=\sin{\left(x \right)}$$$:

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

Alltså,

$$\int{\left(1 - \cot{\left(x \right)}\right)d x} = x - \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)}$$

Lägg till integrationskonstanten:

$$\int{\left(1 - \cot{\left(x \right)}\right)d x} = x - \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)}+C$$

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