Integraal van $$$\tan^{2}{\left(2 x \right)}$$$

Bepaal $$$\int \tan^{2}{\left(2 x \right)}\, dx$$$.

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 kan worden herschreven als

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

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 $$$f{\left(u \right)} = \tan^{2}{\left(u \right)}$$$:

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

Zij $$$v=\tan{\left(u \right)}$$$.

Dan $$$u=\operatorname{atan}{\left(v \right)}$$$ en $$$du=\left(\operatorname{atan}{\left(v \right)}\right)^{\prime }dv = \frac{dv}{v^{2} + 1}$$$ (de stappen zijn te zien »).

De integraal wordt

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

Herschrijf en splits de breuk:

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

Integreer termgewijs:

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

Pas de constantenregel $$$\int c\, dv = c v$$$ toe met $$$c=1$$$:

$$- \frac{\int{\frac{1}{v^{2} + 1} d v}}{2} + \frac{{\color{red}{\int{1 d v}}}}{2} = - \frac{\int{\frac{1}{v^{2} + 1} d v}}{2} + \frac{{\color{red}{v}}}{2}$$

De integraal van $$$\frac{1}{v^{2} + 1}$$$ is $$$\int{\frac{1}{v^{2} + 1} d v} = \operatorname{atan}{\left(v \right)}$$$:

$$\frac{v}{2} - \frac{{\color{red}{\int{\frac{1}{v^{2} + 1} d v}}}}{2} = \frac{v}{2} - \frac{{\color{red}{\operatorname{atan}{\left(v \right)}}}}{2}$$

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

$$- \frac{\operatorname{atan}{\left({\color{red}{v}} \right)}}{2} + \frac{{\color{red}{v}}}{2} = - \frac{\operatorname{atan}{\left({\color{red}{\tan{\left(u \right)}}} \right)}}{2} + \frac{{\color{red}{\tan{\left(u \right)}}}}{2}$$

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

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

Dus,

$$\int{\tan^{2}{\left(2 x \right)} d x} = \frac{\tan{\left(2 x \right)}}{2} - \frac{\operatorname{atan}{\left(\tan{\left(2 x \right)} \right)}}{2}$$

Vereenvoudig:

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

Voeg de integratieconstante toe:

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

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