Integraal van $$$\cos^{3}{\left(x \right)}$$$

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

Haal één cosinus eruit en druk de rest uit in termen van de sinus, met behulp van de formule $$$\cos^2\left(\alpha \right)=-\sin^2\left(\alpha \right)+1$$$ met $$$\alpha=x$$$:

$${\color{red}{\int{\cos^{3}{\left(x \right)} d x}}} = {\color{red}{\int{\left(1 - \sin^{2}{\left(x \right)}\right) \cos{\left(x \right)} d x}}}$$

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

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

De integraal kan worden herschreven als

$${\color{red}{\int{\left(1 - \sin^{2}{\left(x \right)}\right) \cos{\left(x \right)} d x}}} = {\color{red}{\int{\left(1 - u^{2}\right)d u}}}$$

Integreer termgewijs:

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

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 $$$u=\sin{\left(x \right)}$$$:

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

Dus,

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

Voeg de integratieconstante toe:

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

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