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

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

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 $$$\alpha=x$$$:

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

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

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

Dus,

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

Integreer termgewijs:

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

Pas de constantenregel $$$\int c\, du = c u$$$ toe met $$$c=1$$$:

$$\int{u^{2} d u} + {\color{red}{\int{1 d u}}} = \int{u^{2} d u} + {\color{red}{u}}$$

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

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

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

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

Dus,

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

Voeg de integratieconstante toe:

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

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