Integraal van $$$8 \tan{\left(x \right)} \sec^{3}{\left(x \right)}$$$

Bepaal $$$\int 8 \tan{\left(x \right)} \sec^{3}{\left(x \right)}\, dx$$$.

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=8$$$ en $$$f{\left(x \right)} = \tan{\left(x \right)} \sec^{3}{\left(x \right)}$$$:

$${\color{red}{\int{8 \tan{\left(x \right)} \sec^{3}{\left(x \right)} d x}}} = {\color{red}{\left(8 \int{\tan{\left(x \right)} \sec^{3}{\left(x \right)} d x}\right)}}$$

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

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

Dus,

$$8 {\color{red}{\int{\tan{\left(x \right)} \sec^{3}{\left(x \right)} d x}}} = 8 {\color{red}{\int{u^{2} d u}}}$$

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

$$8 {\color{red}{\int{u^{2} d u}}}=8 {\color{red}{\frac{u^{1 + 2}}{1 + 2}}}=8 {\color{red}{\left(\frac{u^{3}}{3}\right)}}$$

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

$$\frac{8 {\color{red}{u}}^{3}}{3} = \frac{8 {\color{red}{\sec{\left(x \right)}}}^{3}}{3}$$

Dus,

$$\int{8 \tan{\left(x \right)} \sec^{3}{\left(x \right)} d x} = \frac{8 \sec^{3}{\left(x \right)}}{3}$$

Voeg de integratieconstante toe:

$$\int{8 \tan{\left(x \right)} \sec^{3}{\left(x \right)} d x} = \frac{8 \sec^{3}{\left(x \right)}}{3}+C$$

$$$\int 8 \tan{\left(x \right)} \sec^{3}{\left(x \right)}\, dx = \frac{8 \sec^{3}{\left(x \right)}}{3} + C$$$A