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

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

Herschrijf in termen van de cotangens:

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

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

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

Dus,

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

Pas de constante-veelvoudregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ toe met $$$c=-1$$$ en $$$f{\left(u \right)} = \frac{u^{4}}{u^{2} + 1}$$$:

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

Aangezien de graad van de teller niet kleiner is dan die van de noemer, voer een staartdeling van polynomen uit (stappen zijn te zien »):

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

Integreer termgewijs:

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

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

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

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

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

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

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

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

$$- \operatorname{atan}{\left({\color{red}{u}} \right)} + {\color{red}{u}} - \frac{{\color{red}{u}}^{3}}{3} = - \operatorname{atan}{\left({\color{red}{\cot{\left(x \right)}}} \right)} + {\color{red}{\cot{\left(x \right)}}} - \frac{{\color{red}{\cot{\left(x \right)}}}^{3}}{3}$$

Dus,

$$\int{\frac{\cos^{4}{\left(x \right)}}{\sin^{4}{\left(x \right)}} d x} = - \frac{\cot^{3}{\left(x \right)}}{3} + \cot{\left(x \right)} - \operatorname{atan}{\left(\cot{\left(x \right)} \right)}$$

Voeg de integratieconstante toe:

$$\int{\frac{\cos^{4}{\left(x \right)}}{\sin^{4}{\left(x \right)}} d x} = - \frac{\cot^{3}{\left(x \right)}}{3} + \cot{\left(x \right)} - \operatorname{atan}{\left(\cot{\left(x \right)} \right)}+C$$

$$$\int \frac{\cos^{4}{\left(x \right)}}{\sin^{4}{\left(x \right)}}\, dx = \left(- \frac{\cot^{3}{\left(x \right)}}{3} + \cot{\left(x \right)} - \operatorname{atan}{\left(\cot{\left(x \right)} \right)}\right) + C$$$A