Integraal van $$$\sin^{3}{\left(x \right)} + 1$$$

Bepaal $$$\int \left(\sin^{3}{\left(x \right)} + 1\right)\, dx$$$.

Integreer termgewijs:

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

Pas de constantenregel $$$\int c\, dx = c x$$$ toe met $$$c=1$$$:

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

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

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

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

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

Dus,

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

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

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

Integreer termgewijs:

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

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

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

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

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

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

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

Dus,

$$\int{\left(\sin^{3}{\left(x \right)} + 1\right)d x} = x + \frac{\cos^{3}{\left(x \right)}}{3} - \cos{\left(x \right)}$$

Voeg de integratieconstante toe:

$$\int{\left(\sin^{3}{\left(x \right)} + 1\right)d x} = x + \frac{\cos^{3}{\left(x \right)}}{3} - \cos{\left(x \right)}+C$$

$$$\int \left(\sin^{3}{\left(x \right)} + 1\right)\, dx = \left(x + \frac{\cos^{3}{\left(x \right)}}{3} - \cos{\left(x \right)}\right) + C$$$A