Integral von $$$t \sin^{2}{\left(\omega \right)}$$$ nach $$$t$$$

Bestimme $$$\int t \sin^{2}{\left(\omega \right)}\, dt$$$.

Wende die Konstantenfaktorregel $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ mit $$$c=\sin^{2}{\left(\omega \right)}$$$ und $$$f{\left(t \right)} = t$$$ an:

$${\color{red}{\int{t \sin^{2}{\left(\omega \right)} d t}}} = {\color{red}{\sin^{2}{\left(\omega \right)} \int{t d t}}}$$

Wenden Sie die Potenzregel $$$\int t^{n}\, dt = \frac{t^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=1$$$ an:

$$\sin^{2}{\left(\omega \right)} {\color{red}{\int{t d t}}}=\sin^{2}{\left(\omega \right)} {\color{red}{\frac{t^{1 + 1}}{1 + 1}}}=\sin^{2}{\left(\omega \right)} {\color{red}{\left(\frac{t^{2}}{2}\right)}}$$

Daher,

$$\int{t \sin^{2}{\left(\omega \right)} d t} = \frac{t^{2} \sin^{2}{\left(\omega \right)}}{2}$$

Fügen Sie die Integrationskonstante hinzu:

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

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