Integralen av $$$\frac{1}{2 \sin{\left(\frac{x}{2} \right)}}$$$

Bestäm $$$\int \frac{1}{2 \sin{\left(\frac{x}{2} \right)}}\, dx$$$.

Skriv om sinus med hjälp av dubbelvinkelformeln $$$\sin\left(\frac{x}{2}\right)=2\sin\left(\frac{\frac{x}{2}}{2}\right)\cos\left(\frac{\frac{x}{2}}{2}\right)$$$:

$${\color{red}{\int{\frac{1}{2 \sin{\left(\frac{x}{2} \right)}} d x}}} = {\color{red}{\int{\frac{1}{4 \sin{\left(\frac{x}{4} \right)} \cos{\left(\frac{x}{4} \right)}} d x}}}$$

Multiplicera täljare och nämnare med $$$\sec^2\left(\frac{x}{4} \right)$$$:

$${\color{red}{\int{\frac{1}{4 \sin{\left(\frac{x}{4} \right)} \cos{\left(\frac{x}{4} \right)}} d x}}} = {\color{red}{\int{\frac{\sec^{2}{\left(\frac{x}{4} \right)}}{4 \tan{\left(\frac{x}{4} \right)}} d x}}}$$

Låt $$$u=\tan{\left(\frac{x}{4} \right)}$$$ vara.

Då $$$du=\left(\tan{\left(\frac{x}{4} \right)}\right)^{\prime }dx = \frac{\sec^{2}{\left(\frac{x}{4} \right)}}{4} dx$$$ (stegen kan ses »), och vi har att $$$\sec^{2}{\left(\frac{x}{4} \right)} dx = 4 du$$$.

Alltså,

$${\color{red}{\int{\frac{\sec^{2}{\left(\frac{x}{4} \right)}}{4 \tan{\left(\frac{x}{4} \right)}} d x}}} = {\color{red}{\int{\frac{1}{u} d u}}}$$

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 $$$u=\tan{\left(\frac{x}{4} \right)}$$$:

$$\ln{\left(\left|{{\color{red}{u}}}\right| \right)} = \ln{\left(\left|{{\color{red}{\tan{\left(\frac{x}{4} \right)}}}}\right| \right)}$$

Alltså,

$$\int{\frac{1}{2 \sin{\left(\frac{x}{2} \right)}} d x} = \ln{\left(\left|{\tan{\left(\frac{x}{4} \right)}}\right| \right)}$$

Lägg till integrationskonstanten:

$$\int{\frac{1}{2 \sin{\left(\frac{x}{2} \right)}} d x} = \ln{\left(\left|{\tan{\left(\frac{x}{4} \right)}}\right| \right)}+C$$

$$$\int \frac{1}{2 \sin{\left(\frac{x}{2} \right)}}\, dx = \ln\left(\left|{\tan{\left(\frac{x}{4} \right)}}\right|\right) + C$$$A