Integral von $$$\frac{b^{2} \sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)}}{a^{2}}$$$ nach $$$x$$$
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Ihre Eingabe
Bestimme $$$\int \frac{b^{2} \sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)}}{a^{2}}\, dx$$$.
Lösung
Schreiben Sie den Integranden um:
$${\color{red}{\int{\frac{b^{2} \sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)}}{a^{2}} d x}}} = {\color{red}{\int{\frac{2 b^{2} \sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}}{a^{2}} d x}}}$$
Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=\frac{2 b^{2}}{a^{2}}$$$ und $$$f{\left(x \right)} = \sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}$$$ an:
$${\color{red}{\int{\frac{2 b^{2} \sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}}{a^{2}} d x}}} = {\color{red}{\left(\frac{2 b^{2} \int{\sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)} d x}}{a^{2}}\right)}}$$
Schreibe den Integranden mithilfe der Doppelwinkel-Formel $$$\sin\left(x \right)\cos\left(x \right)=\frac{1}{2}\sin\left( 2 x \right)$$$ um:
$$\frac{2 b^{2} {\color{red}{\int{\sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)} d x}}}}{a^{2}} = \frac{2 b^{2} {\color{red}{\int{\frac{\sin^{2}{\left(2 x \right)}}{4} d x}}}}{a^{2}}$$
Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=\frac{1}{4}$$$ und $$$f{\left(x \right)} = \sin^{2}{\left(2 x \right)}$$$ an:
$$\frac{2 b^{2} {\color{red}{\int{\frac{\sin^{2}{\left(2 x \right)}}{4} d x}}}}{a^{2}} = \frac{2 b^{2} {\color{red}{\left(\frac{\int{\sin^{2}{\left(2 x \right)} d x}}{4}\right)}}}{a^{2}}$$
Wende die Potenzreduktionsformel $$$\sin^{2}{\left(\alpha \right)} = \frac{1}{2} - \frac{\cos{\left(2 \alpha \right)}}{2}$$$ mit $$$\alpha=2 x$$$ an:
$$\frac{b^{2} {\color{red}{\int{\sin^{2}{\left(2 x \right)} d x}}}}{2 a^{2}} = \frac{b^{2} {\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(4 x \right)}}{2}\right)d x}}}}{2 a^{2}}$$
Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=\frac{1}{2}$$$ und $$$f{\left(x \right)} = 1 - \cos{\left(4 x \right)}$$$ an:
$$\frac{b^{2} {\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(4 x \right)}}{2}\right)d x}}}}{2 a^{2}} = \frac{b^{2} {\color{red}{\left(\frac{\int{\left(1 - \cos{\left(4 x \right)}\right)d x}}{2}\right)}}}{2 a^{2}}$$
Gliedweise integrieren:
$$\frac{b^{2} {\color{red}{\int{\left(1 - \cos{\left(4 x \right)}\right)d x}}}}{4 a^{2}} = \frac{b^{2} {\color{red}{\left(\int{1 d x} - \int{\cos{\left(4 x \right)} d x}\right)}}}{4 a^{2}}$$
Wenden Sie die Konstantenregel $$$\int c\, dx = c x$$$ mit $$$c=1$$$ an:
$$\frac{b^{2} \left(- \int{\cos{\left(4 x \right)} d x} + {\color{red}{\int{1 d x}}}\right)}{4 a^{2}} = \frac{b^{2} \left(- \int{\cos{\left(4 x \right)} d x} + {\color{red}{x}}\right)}{4 a^{2}}$$
Sei $$$u=4 x$$$.
Dann $$$du=\left(4 x\right)^{\prime }dx = 4 dx$$$ (die Schritte sind » zu sehen), und es gilt $$$dx = \frac{du}{4}$$$.
Somit,
$$\frac{b^{2} \left(x - {\color{red}{\int{\cos{\left(4 x \right)} d x}}}\right)}{4 a^{2}} = \frac{b^{2} \left(x - {\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}\right)}{4 a^{2}}$$
Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=\frac{1}{4}$$$ und $$$f{\left(u \right)} = \cos{\left(u \right)}$$$ an:
$$\frac{b^{2} \left(x - {\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}\right)}{4 a^{2}} = \frac{b^{2} \left(x - {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{4}\right)}}\right)}{4 a^{2}}$$
Das Integral des Kosinus ist $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:
$$\frac{b^{2} \left(x - \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{4}\right)}{4 a^{2}} = \frac{b^{2} \left(x - \frac{{\color{red}{\sin{\left(u \right)}}}}{4}\right)}{4 a^{2}}$$
Zur Erinnerung: $$$u=4 x$$$:
$$\frac{b^{2} \left(x - \frac{\sin{\left({\color{red}{u}} \right)}}{4}\right)}{4 a^{2}} = \frac{b^{2} \left(x - \frac{\sin{\left({\color{red}{\left(4 x\right)}} \right)}}{4}\right)}{4 a^{2}}$$
Daher,
$$\int{\frac{b^{2} \sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)}}{a^{2}} d x} = \frac{b^{2} \left(x - \frac{\sin{\left(4 x \right)}}{4}\right)}{4 a^{2}}$$
Vereinfachen:
$$\int{\frac{b^{2} \sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)}}{a^{2}} d x} = \frac{b^{2} \left(4 x - \sin{\left(4 x \right)}\right)}{16 a^{2}}$$
Fügen Sie die Integrationskonstante hinzu:
$$\int{\frac{b^{2} \sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)}}{a^{2}} d x} = \frac{b^{2} \left(4 x - \sin{\left(4 x \right)}\right)}{16 a^{2}}+C$$
Antwort
$$$\int \frac{b^{2} \sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)}}{a^{2}}\, dx = \frac{b^{2} \left(4 x - \sin{\left(4 x \right)}\right)}{16 a^{2}} + C$$$A