Integraal van $$$\frac{t \cos{\left(2 t \right)}}{4}$$$

Bepaal $$$\int \frac{t \cos{\left(2 t \right)}}{4}\, dt$$$.

Pas de constante-veelvoudregel $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ toe met $$$c=\frac{1}{4}$$$ en $$$f{\left(t \right)} = t \cos{\left(2 t \right)}$$$:

$${\color{red}{\int{\frac{t \cos{\left(2 t \right)}}{4} d t}}} = {\color{red}{\left(\frac{\int{t \cos{\left(2 t \right)} d t}}{4}\right)}}$$

Voor de integraal $$$\int{t \cos{\left(2 t \right)} d t}$$$, gebruik partiële integratie $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Zij $$$\operatorname{u}=t$$$ en $$$\operatorname{dv}=\cos{\left(2 t \right)} dt$$$.

Dan $$$\operatorname{du}=\left(t\right)^{\prime }dt=1 dt$$$ (de stappen zijn te zien ») en $$$\operatorname{v}=\int{\cos{\left(2 t \right)} d t}=\frac{\sin{\left(2 t \right)}}{2}$$$ (de stappen zijn te zien »).

Dus,

$$\frac{{\color{red}{\int{t \cos{\left(2 t \right)} d t}}}}{4}=\frac{{\color{red}{\left(t \cdot \frac{\sin{\left(2 t \right)}}{2}-\int{\frac{\sin{\left(2 t \right)}}{2} \cdot 1 d t}\right)}}}{4}=\frac{{\color{red}{\left(\frac{t \sin{\left(2 t \right)}}{2} - \int{\frac{\sin{\left(2 t \right)}}{2} d t}\right)}}}{4}$$

Pas de constante-veelvoudregel $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ toe met $$$c=\frac{1}{2}$$$ en $$$f{\left(t \right)} = \sin{\left(2 t \right)}$$$:

$$\frac{t \sin{\left(2 t \right)}}{8} - \frac{{\color{red}{\int{\frac{\sin{\left(2 t \right)}}{2} d t}}}}{4} = \frac{t \sin{\left(2 t \right)}}{8} - \frac{{\color{red}{\left(\frac{\int{\sin{\left(2 t \right)} d t}}{2}\right)}}}{4}$$

Zij $$$u=2 t$$$.

Dan $$$du=\left(2 t\right)^{\prime }dt = 2 dt$$$ (de stappen zijn te zien »), en dan geldt dat $$$dt = \frac{du}{2}$$$.

Dus,

$$\frac{t \sin{\left(2 t \right)}}{8} - \frac{{\color{red}{\int{\sin{\left(2 t \right)} d t}}}}{8} = \frac{t \sin{\left(2 t \right)}}{8} - \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}}}{8}$$

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

$$\frac{t \sin{\left(2 t \right)}}{8} - \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}}}{8} = \frac{t \sin{\left(2 t \right)}}{8} - \frac{{\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{2}\right)}}}{8}$$

De integraal van de sinus is $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$:

$$\frac{t \sin{\left(2 t \right)}}{8} - \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{16} = \frac{t \sin{\left(2 t \right)}}{8} - \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{16}$$

We herinneren eraan dat $$$u=2 t$$$:

$$\frac{t \sin{\left(2 t \right)}}{8} + \frac{\cos{\left({\color{red}{u}} \right)}}{16} = \frac{t \sin{\left(2 t \right)}}{8} + \frac{\cos{\left({\color{red}{\left(2 t\right)}} \right)}}{16}$$

Dus,

$$\int{\frac{t \cos{\left(2 t \right)}}{4} d t} = \frac{t \sin{\left(2 t \right)}}{8} + \frac{\cos{\left(2 t \right)}}{16}$$

Voeg de integratieconstante toe:

$$\int{\frac{t \cos{\left(2 t \right)}}{4} d t} = \frac{t \sin{\left(2 t \right)}}{8} + \frac{\cos{\left(2 t \right)}}{16}+C$$

$$$\int \frac{t \cos{\left(2 t \right)}}{4}\, dt = \left(\frac{t \sin{\left(2 t \right)}}{8} + \frac{\cos{\left(2 t \right)}}{16}\right) + C$$$A