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

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

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

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

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

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

Dus,

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

Herschrijf de integraand in termen van secans:

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

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

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

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

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

Dus,

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

Voeg de integratieconstante toe:

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

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