Integraal van $$$\tan^{2}{\left(3 x \right)} \sec^{4}{\left(3 x \right)}$$$

Bepaal $$$\int \tan^{2}{\left(3 x \right)} \sec^{4}{\left(3 x \right)}\, dx$$$.

Haal twee secansen eruit en schrijf de rest in termen van de tangens, met behulp van de formule $$$\sec^2\left( \alpha \right)=\tan^2\left( \alpha \right) + 1$$$ met $$$\alpha=3 x$$$:

$${\color{red}{\int{\tan^{2}{\left(3 x \right)} \sec^{4}{\left(3 x \right)} d x}}} = {\color{red}{\int{\left(\tan^{2}{\left(3 x \right)} + 1\right) \tan^{2}{\left(3 x \right)} \sec^{2}{\left(3 x \right)} d x}}}$$

Zij $$$u=\tan{\left(3 x \right)}$$$.

Dan $$$du=\left(\tan{\left(3 x \right)}\right)^{\prime }dx = 3 \sec^{2}{\left(3 x \right)} dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$\sec^{2}{\left(3 x \right)} dx = \frac{du}{3}$$$.

Dus,

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

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 $$$f{\left(u \right)} = u^{2} \left(u^{2} + 1\right)$$$:

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

Expand the expression:

$$\frac{{\color{red}{\int{u^{2} \left(u^{2} + 1\right) d u}}}}{3} = \frac{{\color{red}{\int{\left(u^{4} + u^{2}\right)d u}}}}{3}$$

Integreer termgewijs:

$$\frac{{\color{red}{\int{\left(u^{4} + u^{2}\right)d u}}}}{3} = \frac{{\color{red}{\left(\int{u^{2} d u} + \int{u^{4} d u}\right)}}}{3}$$

Pas de machtsregel $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ toe met $$$n=2$$$:

$$\frac{\int{u^{4} d u}}{3} + \frac{{\color{red}{\int{u^{2} d u}}}}{3}=\frac{\int{u^{4} d u}}{3} + \frac{{\color{red}{\frac{u^{1 + 2}}{1 + 2}}}}{3}=\frac{\int{u^{4} d u}}{3} + \frac{{\color{red}{\left(\frac{u^{3}}{3}\right)}}}{3}$$

Pas de machtsregel $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ toe met $$$n=4$$$:

$$\frac{u^{3}}{9} + \frac{{\color{red}{\int{u^{4} d u}}}}{3}=\frac{u^{3}}{9} + \frac{{\color{red}{\frac{u^{1 + 4}}{1 + 4}}}}{3}=\frac{u^{3}}{9} + \frac{{\color{red}{\left(\frac{u^{5}}{5}\right)}}}{3}$$

We herinneren eraan dat $$$u=\tan{\left(3 x \right)}$$$:

$$\frac{{\color{red}{u}}^{3}}{9} + \frac{{\color{red}{u}}^{5}}{15} = \frac{{\color{red}{\tan{\left(3 x \right)}}}^{3}}{9} + \frac{{\color{red}{\tan{\left(3 x \right)}}}^{5}}{15}$$

Dus,

$$\int{\tan^{2}{\left(3 x \right)} \sec^{4}{\left(3 x \right)} d x} = \frac{\tan^{5}{\left(3 x \right)}}{15} + \frac{\tan^{3}{\left(3 x \right)}}{9}$$

Voeg de integratieconstante toe:

$$\int{\tan^{2}{\left(3 x \right)} \sec^{4}{\left(3 x \right)} d x} = \frac{\tan^{5}{\left(3 x \right)}}{15} + \frac{\tan^{3}{\left(3 x \right)}}{9}+C$$

$$$\int \tan^{2}{\left(3 x \right)} \sec^{4}{\left(3 x \right)}\, dx = \left(\frac{\tan^{5}{\left(3 x \right)}}{15} + \frac{\tan^{3}{\left(3 x \right)}}{9}\right) + C$$$A