Integraal van $$$\frac{\sin{\left(2 x \right)}}{\sin{\left(4 x \right)}}$$$

Bepaal $$$\int \frac{\sin{\left(2 x \right)}}{\sin{\left(4 x \right)}}\, dx$$$.

Zij $$$u=2 x$$$.

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

De integraal wordt

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

Pas de constante-veelvoudregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ toe met $$$c=\frac{1}{2}$$$ en $$$f{\left(u \right)} = \frac{\sin{\left(u \right)}}{\sin{\left(2 u \right)}}$$$:

$${\color{red}{\int{\frac{\sin{\left(u \right)}}{2 \sin{\left(2 u \right)}} d u}}} = {\color{red}{\left(\frac{\int{\frac{\sin{\left(u \right)}}{\sin{\left(2 u \right)}} d u}}{2}\right)}}$$

Herschrijf de integraand:

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

Herschrijf de cosinus in termen van de sinus met behulp van de formule $$$\cos\left( u \right)=\sin\left( u + \frac{\pi}{2}\right)$$$ en herschrijf vervolgens de sinus met behulp van de dubbelhoekformule $$$\sin\left( u \right)=2\sin\left(\frac{ u }{2}\right)\cos\left(\frac{ u }{2}\right)$$$:

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

Vermenigvuldig de teller en de noemer met $$$\sec^2\left(\frac{ u }{2} + \frac{\pi}{4} \right)$$$:

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

Zij $$$v=\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}$$$.

Dan $$$dv=\left(\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}\right)^{\prime }du = \frac{\sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)}}{2} du$$$ (de stappen zijn te zien »), en dan geldt dat $$$\sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)} du = 2 dv$$$.

Dus,

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

Pas de constante-veelvoudregel $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ toe met $$$c=\frac{1}{2}$$$ en $$$f{\left(v \right)} = \frac{1}{v}$$$:

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

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

$$\frac{{\color{red}{\int{\frac{1}{v} d v}}}}{4} = \frac{{\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{4}$$

We herinneren eraan dat $$$v=\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}$$$:

$$\frac{\ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{4} = \frac{\ln{\left(\left|{{\color{red}{\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}}}}\right| \right)}}{4}$$

We herinneren eraan dat $$$u=2 x$$$:

$$\frac{\ln{\left(\left|{\tan{\left(\frac{\pi}{4} + \frac{{\color{red}{u}}}{2} \right)}}\right| \right)}}{4} = \frac{\ln{\left(\left|{\tan{\left(\frac{\pi}{4} + \frac{{\color{red}{\left(2 x\right)}}}{2} \right)}}\right| \right)}}{4}$$

Dus,

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

Voeg de integratieconstante toe:

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

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