Integralen av $$$\frac{\sin{\left(x \right)}}{1 - \cos{\left(x \right)}}$$$

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

Låt $$$u=1 - \cos{\left(x \right)}$$$ vara.

Då $$$du=\left(1 - \cos{\left(x \right)}\right)^{\prime }dx = \sin{\left(x \right)} dx$$$ (stegen kan ses »), och vi har att $$$\sin{\left(x \right)} dx = du$$$.

Alltså,

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

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

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

Kom ihåg att $$$u=1 - \cos{\left(x \right)}$$$:

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

Alltså,

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

Lägg till integrationskonstanten:

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

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