Integraal van $$$\frac{\sin^{4}{\left(x \right)}}{\cos^{6}{\left(x \right)}}$$$

Bepaal $$$\int \frac{\sin^{4}{\left(x \right)}}{\cos^{6}{\left(x \right)}}\, dx$$$.

Vermenigvuldig de teller en de noemer met $$$\cos^{4}{\left(x \right)}$$$ en zet $$$\frac{\sin^{4}{\left(x \right)}}{\cos^{4}{\left(x \right)}}$$$ om in $$$\tan^{4}{\left(x \right)}$$$:

$${\color{red}{\int{\frac{\sin^{4}{\left(x \right)}}{\cos^{6}{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{\tan^{4}{\left(x \right)}}{\cos^{2}{\left(x \right)}} d x}}}$$

Zet $$$\frac{1}{\cos^{2}{\left(x \right)}}$$$ om in $$$\sec^{2}{\left(x \right)}$$$:

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

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,

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

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

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

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

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

Dus,

$$\int{\frac{\sin^{4}{\left(x \right)}}{\cos^{6}{\left(x \right)}} d x} = \frac{\tan^{5}{\left(x \right)}}{5}$$

Voeg de integratieconstante toe:

$$\int{\frac{\sin^{4}{\left(x \right)}}{\cos^{6}{\left(x \right)}} d x} = \frac{\tan^{5}{\left(x \right)}}{5}+C$$

$$$\int \frac{\sin^{4}{\left(x \right)}}{\cos^{6}{\left(x \right)}}\, dx = \frac{\tan^{5}{\left(x \right)}}{5} + C$$$A