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

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

Expand the expression:

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

Integrera termvis:

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

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

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

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

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

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

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

Låt $$$u=\tan{\left(\frac{x}{2} \right)}$$$ vara.

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

Integralen blir

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

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 $$$u=\tan{\left(\frac{x}{2} \right)}$$$:

$$- x + \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = - x + \ln{\left(\left|{{\color{red}{\tan{\left(\frac{x}{2} \right)}}}}\right| \right)}$$

Alltså,

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

Lägg till integrationskonstanten:

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

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