Integral von $$$u \sin^{2}{\left(3 x \right)}$$$ nach $$$x$$$

Bestimme $$$\int u \sin^{2}{\left(3 x \right)}\, dx$$$.

Wende die Potenzreduktionsformel $$$\sin^{2}{\left(\alpha \right)} = \frac{1}{2} - \frac{\cos{\left(2 \alpha \right)}}{2}$$$ mit $$$\alpha=3 x$$$ an:

$${\color{red}{\int{u \sin^{2}{\left(3 x \right)} d x}}} = {\color{red}{\int{\frac{u \left(1 - \cos{\left(6 x \right)}\right)}{2} d x}}}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=\frac{1}{2}$$$ und $$$f{\left(x \right)} = u \left(1 - \cos{\left(6 x \right)}\right)$$$ an:

$${\color{red}{\int{\frac{u \left(1 - \cos{\left(6 x \right)}\right)}{2} d x}}} = {\color{red}{\left(\frac{\int{u \left(1 - \cos{\left(6 x \right)}\right) d x}}{2}\right)}}$$

Expand the expression:

$$\frac{{\color{red}{\int{u \left(1 - \cos{\left(6 x \right)}\right) d x}}}}{2} = \frac{{\color{red}{\int{\left(- u \cos{\left(6 x \right)} + u\right)d x}}}}{2}$$

Gliedweise integrieren:

$$\frac{{\color{red}{\int{\left(- u \cos{\left(6 x \right)} + u\right)d x}}}}{2} = \frac{{\color{red}{\left(\int{u d x} - \int{u \cos{\left(6 x \right)} d x}\right)}}}{2}$$

Wenden Sie die Konstantenregel $$$\int c\, dx = c x$$$ mit $$$c=u$$$ an:

$$- \frac{\int{u \cos{\left(6 x \right)} d x}}{2} + \frac{{\color{red}{\int{u d x}}}}{2} = - \frac{\int{u \cos{\left(6 x \right)} d x}}{2} + \frac{{\color{red}{u x}}}{2}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=u$$$ und $$$f{\left(x \right)} = \cos{\left(6 x \right)}$$$ an:

$$\frac{u x}{2} - \frac{{\color{red}{\int{u \cos{\left(6 x \right)} d x}}}}{2} = \frac{u x}{2} - \frac{{\color{red}{u \int{\cos{\left(6 x \right)} d x}}}}{2}$$

Sei $$$v=6 x$$$.

Dann $$$dv=\left(6 x\right)^{\prime }dx = 6 dx$$$ (die Schritte sind » zu sehen), und es gilt $$$dx = \frac{dv}{6}$$$.

Daher,

$$\frac{u x}{2} - \frac{u {\color{red}{\int{\cos{\left(6 x \right)} d x}}}}{2} = \frac{u x}{2} - \frac{u {\color{red}{\int{\frac{\cos{\left(v \right)}}{6} d v}}}}{2}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ mit $$$c=\frac{1}{6}$$$ und $$$f{\left(v \right)} = \cos{\left(v \right)}$$$ an:

$$\frac{u x}{2} - \frac{u {\color{red}{\int{\frac{\cos{\left(v \right)}}{6} d v}}}}{2} = \frac{u x}{2} - \frac{u {\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{6}\right)}}}{2}$$

Das Integral des Kosinus ist $$$\int{\cos{\left(v \right)} d v} = \sin{\left(v \right)}$$$:

$$\frac{u x}{2} - \frac{u {\color{red}{\int{\cos{\left(v \right)} d v}}}}{12} = \frac{u x}{2} - \frac{u {\color{red}{\sin{\left(v \right)}}}}{12}$$

Zur Erinnerung: $$$v=6 x$$$:

$$\frac{u x}{2} - \frac{u \sin{\left({\color{red}{v}} \right)}}{12} = \frac{u x}{2} - \frac{u \sin{\left({\color{red}{\left(6 x\right)}} \right)}}{12}$$

Daher,

$$\int{u \sin^{2}{\left(3 x \right)} d x} = \frac{u x}{2} - \frac{u \sin{\left(6 x \right)}}{12}$$

Vereinfachen:

$$\int{u \sin^{2}{\left(3 x \right)} d x} = \frac{u \left(6 x - \sin{\left(6 x \right)}\right)}{12}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{u \sin^{2}{\left(3 x \right)} d x} = \frac{u \left(6 x - \sin{\left(6 x \right)}\right)}{12}+C$$

$$$\int u \sin^{2}{\left(3 x \right)}\, dx = \frac{u \left(6 x - \sin{\left(6 x \right)}\right)}{12} + C$$$A