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

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

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 $$$\cos{\left(x \right)} dx = du$$$.

Alltså,

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

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,

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

Tillämpa potensregeln $$$\int v^{n}\, dv = \frac{v^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ med $$$n=1$$$:

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

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

$$\frac{{\color{red}{v}}^{2}}{2} = \frac{{\color{red}{\ln{\left(u \right)}}}^{2}}{2}$$

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

$$\frac{\ln{\left({\color{red}{u}} \right)}^{2}}{2} = \frac{\ln{\left({\color{red}{\sin{\left(x \right)}}} \right)}^{2}}{2}$$

Alltså,

$$\int{\ln{\left(\sin{\left(x \right)} \right)} \cot{\left(x \right)} d x} = \frac{\ln{\left(\sin{\left(x \right)} \right)}^{2}}{2}$$

Lägg till integrationskonstanten:

$$\int{\ln{\left(\sin{\left(x \right)} \right)} \cot{\left(x \right)} d x} = \frac{\ln{\left(\sin{\left(x \right)} \right)}^{2}}{2}+C$$

$$$\int \ln\left(\sin{\left(x \right)}\right) \cot{\left(x \right)}\, dx = \frac{\ln^{2}\left(\sin{\left(x \right)}\right)}{2} + C$$$A