Integraal van $$$\frac{1}{\cos{\left(3 x \right)} + 1}$$$

Bepaal $$$\int \frac{1}{\cos{\left(3 x \right)} + 1}\, dx$$$.

Zij $$$u=3 x$$$.

Dan $$$du=\left(3 x\right)^{\prime }dx = 3 dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$dx = \frac{du}{3}$$$.

Dus,

$${\color{red}{\int{\frac{1}{\cos{\left(3 x \right)} + 1} d x}}} = {\color{red}{\int{\frac{1}{3 \left(\cos{\left(u \right)} + 1\right)} d u}}}$$

Pas de constante-veelvoudregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ toe met $$$c=\frac{1}{3}$$$ en $$$f{\left(u \right)} = \frac{1}{\cos{\left(u \right)} + 1}$$$:

$${\color{red}{\int{\frac{1}{3 \left(\cos{\left(u \right)} + 1\right)} d u}}} = {\color{red}{\left(\frac{\int{\frac{1}{\cos{\left(u \right)} + 1} d u}}{3}\right)}}$$

Herschrijf de cosinus met behulp van de dubbelhoekformule $$$\cos\left( u \right)=2\cos^2\left(\frac{ u }{2}\right)-1$$$ en vereenvoudig:

$$\frac{{\color{red}{\int{\frac{1}{\cos{\left(u \right)} + 1} d u}}}}{3} = \frac{{\color{red}{\int{\frac{1}{2 \cos^{2}{\left(\frac{u}{2} \right)}} d u}}}}{3}$$

Zij $$$v=\frac{u}{2}$$$.

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

Dus,

$$\frac{{\color{red}{\int{\frac{1}{2 \cos^{2}{\left(\frac{u}{2} \right)}} d u}}}}{3} = \frac{{\color{red}{\int{\frac{1}{\cos^{2}{\left(v \right)}} d v}}}}{3}$$

Herschrijf de integraand in termen van secans:

$$\frac{{\color{red}{\int{\frac{1}{\cos^{2}{\left(v \right)}} d v}}}}{3} = \frac{{\color{red}{\int{\sec^{2}{\left(v \right)} d v}}}}{3}$$

De integraal van $$$\sec^{2}{\left(v \right)}$$$ is $$$\int{\sec^{2}{\left(v \right)} d v} = \tan{\left(v \right)}$$$:

$$\frac{{\color{red}{\int{\sec^{2}{\left(v \right)} d v}}}}{3} = \frac{{\color{red}{\tan{\left(v \right)}}}}{3}$$

We herinneren eraan dat $$$v=\frac{u}{2}$$$:

$$\frac{\tan{\left({\color{red}{v}} \right)}}{3} = \frac{\tan{\left({\color{red}{\left(\frac{u}{2}\right)}} \right)}}{3}$$

We herinneren eraan dat $$$u=3 x$$$:

$$\frac{\tan{\left(\frac{{\color{red}{u}}}{2} \right)}}{3} = \frac{\tan{\left(\frac{{\color{red}{\left(3 x\right)}}}{2} \right)}}{3}$$

Dus,

$$\int{\frac{1}{\cos{\left(3 x \right)} + 1} d x} = \frac{\tan{\left(\frac{3 x}{2} \right)}}{3}$$

Voeg de integratieconstante toe:

$$\int{\frac{1}{\cos{\left(3 x \right)} + 1} d x} = \frac{\tan{\left(\frac{3 x}{2} \right)}}{3}+C$$

$$$\int \frac{1}{\cos{\left(3 x \right)} + 1}\, dx = \frac{\tan{\left(\frac{3 x}{2} \right)}}{3} + C$$$A