Integraal van $$$\frac{\cos{\left(2 x \right)}}{\sin{\left(x \right)}}$$$

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

Herschrijf de cosinus met behulp van de dubbelhoekformule $$$\cos\left(2x\right)=-2\sin^2\left(x\right)+1$$$:

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

Splits de breuk:

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

Integreer termgewijs:

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

Herschrijf de sinus met behulp van de formule voor de dubbele hoek $$$\sin\left(x\right)=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)$$$:

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

Vermenigvuldig de teller en de noemer met $$$\sec^2\left(\frac{x}{2} \right)$$$:

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

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

Dan $$$du=\left(\tan{\left(\frac{x}{2} \right)}\right)^{\prime }dx = \frac{\sec^{2}{\left(\frac{x}{2} \right)}}{2} dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$\sec^{2}{\left(\frac{x}{2} \right)} dx = 2 du$$$.

De integraal wordt

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

De integraal van $$$\frac{1}{u}$$$ is $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

We herinneren eraan dat $$$u=\tan{\left(\frac{x}{2} \right)}$$$:

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

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=2$$$ en $$$f{\left(x \right)} = \sin{\left(x \right)}$$$:

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

De integraal van de sinus is $$$\int{\sin{\left(x \right)} d x} = - \cos{\left(x \right)}$$$:

$$\ln{\left(\left|{\tan{\left(\frac{x}{2} \right)}}\right| \right)} - 2 {\color{red}{\int{\sin{\left(x \right)} d x}}} = \ln{\left(\left|{\tan{\left(\frac{x}{2} \right)}}\right| \right)} - 2 {\color{red}{\left(- \cos{\left(x \right)}\right)}}$$

Dus,

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

Voeg de integratieconstante toe:

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

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