Integralen av $$$\frac{\cot{\left(x \right)}}{\ln\left(\sin{\left(x \right)}\right)}$$$

Bestäm $$$\int \frac{\cot{\left(x \right)}}{\ln\left(\sin{\left(x \right)}\right)}\, dx$$$.

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 kan omskrivas som

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

Låt $$$v=\ln{\left(u \right)}$$$ vara.

Då $$$dv=\left(\ln{\left(u \right)}\right)^{\prime }du = \frac{du}{u}$$$ (stegen kan ses »), och vi har att $$$\frac{du}{u} = dv$$$.

Alltså,

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

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

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

Kom ihåg att $$$v=\ln{\left(u \right)}$$$:

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

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

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

Alltså,

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

Lägg till integrationskonstanten:

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

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