Integraal van $$$\left(\cos{\left(t \right)} + 1\right) \sin{\left(t \right)}$$$

Bepaal $$$\int \left(\cos{\left(t \right)} + 1\right) \sin{\left(t \right)}\, dt$$$.

Zij $$$u=\cos{\left(t \right)} + 1$$$.

Dan $$$du=\left(\cos{\left(t \right)} + 1\right)^{\prime }dt = - \sin{\left(t \right)} dt$$$ (de stappen zijn te zien »), en dan geldt dat $$$\sin{\left(t \right)} dt = - du$$$.

De integraal kan worden herschreven als

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

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

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

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

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

We herinneren eraan dat $$$u=\cos{\left(t \right)} + 1$$$:

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

Dus,

$$\int{\left(\cos{\left(t \right)} + 1\right) \sin{\left(t \right)} d t} = - \frac{\left(\cos{\left(t \right)} + 1\right)^{2}}{2}$$

Voeg de integratieconstante toe:

$$\int{\left(\cos{\left(t \right)} + 1\right) \sin{\left(t \right)} d t} = - \frac{\left(\cos{\left(t \right)} + 1\right)^{2}}{2}+C$$

$$$\int \left(\cos{\left(t \right)} + 1\right) \sin{\left(t \right)}\, dt = - \frac{\left(\cos{\left(t \right)} + 1\right)^{2}}{2} + C$$$A