Integral von $$$i d n t \sin{\left(2 x \right)}$$$ nach $$$x$$$

Bestimme $$$\int i d n t \sin{\left(2 x \right)}\, dx$$$.

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=i d n t$$$ und $$$f{\left(x \right)} = \sin{\left(2 x \right)}$$$ an:

$${\color{red}{\int{i d n t \sin{\left(2 x \right)} d x}}} = {\color{red}{i d n t \int{\sin{\left(2 x \right)} d x}}}$$

Sei $$$u=2 x$$$.

Dann $$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (die Schritte sind » zu sehen), und es gilt $$$dx = \frac{du}{2}$$$.

Daher,

$$i d n t {\color{red}{\int{\sin{\left(2 x \right)} d x}}} = i d n t {\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=\frac{1}{2}$$$ und $$$f{\left(u \right)} = \sin{\left(u \right)}$$$ an:

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

Das Integral des Sinus lautet $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$:

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

Zur Erinnerung: $$$u=2 x$$$:

$$- \frac{i d n t \cos{\left({\color{red}{u}} \right)}}{2} = - \frac{i d n t \cos{\left({\color{red}{\left(2 x\right)}} \right)}}{2}$$

Daher,

$$\int{i d n t \sin{\left(2 x \right)} d x} = - \frac{i d n t \cos{\left(2 x \right)}}{2}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{i d n t \sin{\left(2 x \right)} d x} = - \frac{i d n t \cos{\left(2 x \right)}}{2}+C$$

$$$\int i d n t \sin{\left(2 x \right)}\, dx = - \frac{i d n t \cos{\left(2 x \right)}}{2} + C$$$A