Integral von $$$4 \sin^{2}{\left(x \right)} \cos^{3}{\left(x \right)}$$$

Bestimme $$$\int 4 \sin^{2}{\left(x \right)} \cos^{3}{\left(x \right)}\, dx$$$.

Wende die Potenzreduktionsformel $$$\cos^{3}{\left(\alpha \right)} = \frac{3 \cos{\left(\alpha \right)}}{4} + \frac{\cos{\left(3 \alpha \right)}}{4}$$$ mit $$$\alpha=x$$$ an:

$${\color{red}{\int{4 \sin^{2}{\left(x \right)} \cos^{3}{\left(x \right)} d x}}} = {\color{red}{\int{\left(3 \cos{\left(x \right)} + \cos{\left(3 x \right)}\right) \sin^{2}{\left(x \right)} d x}}}$$

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

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

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=\frac{1}{8}$$$ und $$$f{\left(x \right)} = 4 \left(1 - \cos{\left(2 x \right)}\right) \left(3 \cos{\left(x \right)} + \cos{\left(3 x \right)}\right)$$$ an:

$${\color{red}{\int{\frac{\left(1 - \cos{\left(2 x \right)}\right) \left(3 \cos{\left(x \right)} + \cos{\left(3 x \right)}\right)}{2} d x}}} = {\color{red}{\left(\frac{\int{4 \left(1 - \cos{\left(2 x \right)}\right) \left(3 \cos{\left(x \right)} + \cos{\left(3 x \right)}\right) d x}}{8}\right)}}$$

Expand the expression:

$$\frac{{\color{red}{\int{4 \left(1 - \cos{\left(2 x \right)}\right) \left(3 \cos{\left(x \right)} + \cos{\left(3 x \right)}\right) d x}}}}{8} = \frac{{\color{red}{\int{\left(- 12 \cos{\left(x \right)} \cos{\left(2 x \right)} + 12 \cos{\left(x \right)} - 4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} + 4 \cos{\left(3 x \right)}\right)d x}}}}{8}$$

Gliedweise integrieren:

$$\frac{{\color{red}{\int{\left(- 12 \cos{\left(x \right)} \cos{\left(2 x \right)} + 12 \cos{\left(x \right)} - 4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} + 4 \cos{\left(3 x \right)}\right)d x}}}}{8} = \frac{{\color{red}{\left(- \int{12 \cos{\left(x \right)} \cos{\left(2 x \right)} d x} - \int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x} + \int{12 \cos{\left(x \right)} d x} + \int{4 \cos{\left(3 x \right)} d x}\right)}}}{8}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=4$$$ und $$$f{\left(x \right)} = \cos{\left(3 x \right)}$$$ an:

$$- \frac{\int{12 \cos{\left(x \right)} \cos{\left(2 x \right)} d x}}{8} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} + \frac{\int{12 \cos{\left(x \right)} d x}}{8} + \frac{{\color{red}{\int{4 \cos{\left(3 x \right)} d x}}}}{8} = - \frac{\int{12 \cos{\left(x \right)} \cos{\left(2 x \right)} d x}}{8} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} + \frac{\int{12 \cos{\left(x \right)} d x}}{8} + \frac{{\color{red}{\left(4 \int{\cos{\left(3 x \right)} d x}\right)}}}{8}$$

Sei $$$u=3 x$$$.

Dann $$$du=\left(3 x\right)^{\prime }dx = 3 dx$$$ (die Schritte sind » zu sehen), und es gilt $$$dx = \frac{du}{3}$$$.

Daher,

$$- \frac{\int{12 \cos{\left(x \right)} \cos{\left(2 x \right)} d x}}{8} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} + \frac{\int{12 \cos{\left(x \right)} d x}}{8} + \frac{{\color{red}{\int{\cos{\left(3 x \right)} d x}}}}{2} = - \frac{\int{12 \cos{\left(x \right)} \cos{\left(2 x \right)} d x}}{8} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} + \frac{\int{12 \cos{\left(x \right)} d x}}{8} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{3} d u}}}}{2}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=\frac{1}{3}$$$ und $$$f{\left(u \right)} = \cos{\left(u \right)}$$$ an:

$$- \frac{\int{12 \cos{\left(x \right)} \cos{\left(2 x \right)} d x}}{8} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} + \frac{\int{12 \cos{\left(x \right)} d x}}{8} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{3} d u}}}}{2} = - \frac{\int{12 \cos{\left(x \right)} \cos{\left(2 x \right)} d x}}{8} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} + \frac{\int{12 \cos{\left(x \right)} d x}}{8} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{3}\right)}}}{2}$$

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

$$- \frac{\int{12 \cos{\left(x \right)} \cos{\left(2 x \right)} d x}}{8} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} + \frac{\int{12 \cos{\left(x \right)} d x}}{8} + \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{6} = - \frac{\int{12 \cos{\left(x \right)} \cos{\left(2 x \right)} d x}}{8} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} + \frac{\int{12 \cos{\left(x \right)} d x}}{8} + \frac{{\color{red}{\sin{\left(u \right)}}}}{6}$$

Zur Erinnerung: $$$u=3 x$$$:

$$- \frac{\int{12 \cos{\left(x \right)} \cos{\left(2 x \right)} d x}}{8} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} + \frac{\int{12 \cos{\left(x \right)} d x}}{8} + \frac{\sin{\left({\color{red}{u}} \right)}}{6} = - \frac{\int{12 \cos{\left(x \right)} \cos{\left(2 x \right)} d x}}{8} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} + \frac{\int{12 \cos{\left(x \right)} d x}}{8} + \frac{\sin{\left({\color{red}{\left(3 x\right)}} \right)}}{6}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=12$$$ und $$$f{\left(x \right)} = \cos{\left(x \right)}$$$ an:

$$\frac{\sin{\left(3 x \right)}}{6} - \frac{\int{12 \cos{\left(x \right)} \cos{\left(2 x \right)} d x}}{8} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} + \frac{{\color{red}{\int{12 \cos{\left(x \right)} d x}}}}{8} = \frac{\sin{\left(3 x \right)}}{6} - \frac{\int{12 \cos{\left(x \right)} \cos{\left(2 x \right)} d x}}{8} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} + \frac{{\color{red}{\left(12 \int{\cos{\left(x \right)} d x}\right)}}}{8}$$

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

$$\frac{\sin{\left(3 x \right)}}{6} - \frac{\int{12 \cos{\left(x \right)} \cos{\left(2 x \right)} d x}}{8} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} + \frac{3 {\color{red}{\int{\cos{\left(x \right)} d x}}}}{2} = \frac{\sin{\left(3 x \right)}}{6} - \frac{\int{12 \cos{\left(x \right)} \cos{\left(2 x \right)} d x}}{8} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} + \frac{3 {\color{red}{\sin{\left(x \right)}}}}{2}$$

Schreibe $$$\cos\left(x \right)\cos\left(2 x \right)$$$ mithilfe der Formel $$$\cos\left(\alpha \right)\cos\left(\beta \right)=\frac{1}{2} \cos\left(\alpha-\beta \right)+\frac{1}{2} \cos\left(\alpha+\beta \right)$$$ mit $$$\alpha=x$$$ und $$$\beta=2 x$$$ um:

$$\frac{3 \sin{\left(x \right)}}{2} + \frac{\sin{\left(3 x \right)}}{6} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} - \frac{{\color{red}{\int{12 \cos{\left(x \right)} \cos{\left(2 x \right)} d x}}}}{8} = \frac{3 \sin{\left(x \right)}}{2} + \frac{\sin{\left(3 x \right)}}{6} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} - \frac{{\color{red}{\int{\left(6 \cos{\left(x \right)} + 6 \cos{\left(3 x \right)}\right)d x}}}}{8}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=\frac{1}{2}$$$ und $$$f{\left(x \right)} = 12 \cos{\left(x \right)} + 12 \cos{\left(3 x \right)}$$$ an:

$$\frac{3 \sin{\left(x \right)}}{2} + \frac{\sin{\left(3 x \right)}}{6} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} - \frac{{\color{red}{\int{\left(6 \cos{\left(x \right)} + 6 \cos{\left(3 x \right)}\right)d x}}}}{8} = \frac{3 \sin{\left(x \right)}}{2} + \frac{\sin{\left(3 x \right)}}{6} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} - \frac{{\color{red}{\left(\frac{\int{\left(12 \cos{\left(x \right)} + 12 \cos{\left(3 x \right)}\right)d x}}{2}\right)}}}{8}$$

Gliedweise integrieren:

$$\frac{3 \sin{\left(x \right)}}{2} + \frac{\sin{\left(3 x \right)}}{6} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} - \frac{{\color{red}{\int{\left(12 \cos{\left(x \right)} + 12 \cos{\left(3 x \right)}\right)d x}}}}{16} = \frac{3 \sin{\left(x \right)}}{2} + \frac{\sin{\left(3 x \right)}}{6} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} - \frac{{\color{red}{\left(\int{12 \cos{\left(x \right)} d x} + \int{12 \cos{\left(3 x \right)} d x}\right)}}}{16}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=12$$$ und $$$f{\left(x \right)} = \cos{\left(x \right)}$$$ an:

$$\frac{3 \sin{\left(x \right)}}{2} + \frac{\sin{\left(3 x \right)}}{6} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} - \frac{\int{12 \cos{\left(3 x \right)} d x}}{16} - \frac{{\color{red}{\int{12 \cos{\left(x \right)} d x}}}}{16} = \frac{3 \sin{\left(x \right)}}{2} + \frac{\sin{\left(3 x \right)}}{6} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} - \frac{\int{12 \cos{\left(3 x \right)} d x}}{16} - \frac{{\color{red}{\left(12 \int{\cos{\left(x \right)} d x}\right)}}}{16}$$

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

$$\frac{3 \sin{\left(x \right)}}{2} + \frac{\sin{\left(3 x \right)}}{6} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} - \frac{\int{12 \cos{\left(3 x \right)} d x}}{16} - \frac{3 {\color{red}{\int{\cos{\left(x \right)} d x}}}}{4} = \frac{3 \sin{\left(x \right)}}{2} + \frac{\sin{\left(3 x \right)}}{6} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} - \frac{\int{12 \cos{\left(3 x \right)} d x}}{16} - \frac{3 {\color{red}{\sin{\left(x \right)}}}}{4}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=12$$$ und $$$f{\left(x \right)} = \cos{\left(3 x \right)}$$$ an:

$$\frac{3 \sin{\left(x \right)}}{4} + \frac{\sin{\left(3 x \right)}}{6} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} - \frac{{\color{red}{\int{12 \cos{\left(3 x \right)} d x}}}}{16} = \frac{3 \sin{\left(x \right)}}{4} + \frac{\sin{\left(3 x \right)}}{6} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} - \frac{{\color{red}{\left(12 \int{\cos{\left(3 x \right)} d x}\right)}}}{16}$$

Das Integral $$$\int{\cos{\left(3 x \right)} d x}$$$ wurde bereits berechnet:

$$\int{\cos{\left(3 x \right)} d x} = \frac{\sin{\left(3 x \right)}}{3}$$

Daher,

$$\frac{3 \sin{\left(x \right)}}{4} + \frac{\sin{\left(3 x \right)}}{6} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} - \frac{3 {\color{red}{\int{\cos{\left(3 x \right)} d x}}}}{4} = \frac{3 \sin{\left(x \right)}}{4} + \frac{\sin{\left(3 x \right)}}{6} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} - \frac{3 {\color{red}{\left(\frac{\sin{\left(3 x \right)}}{3}\right)}}}{4}$$

Schreibe $$$\cos\left(2 x \right)\cos\left(3 x \right)$$$ mithilfe der Formel $$$\cos\left(\alpha \right)\cos\left(\beta \right)=\frac{1}{2} \cos\left(\alpha-\beta \right)+\frac{1}{2} \cos\left(\alpha+\beta \right)$$$ mit $$$\alpha=2 x$$$ und $$$\beta=3 x$$$ um:

$$\frac{3 \sin{\left(x \right)}}{4} - \frac{\sin{\left(3 x \right)}}{12} - \frac{{\color{red}{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}}}{8} = \frac{3 \sin{\left(x \right)}}{4} - \frac{\sin{\left(3 x \right)}}{12} - \frac{{\color{red}{\int{\left(2 \cos{\left(x \right)} + 2 \cos{\left(5 x \right)}\right)d x}}}}{8}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=\frac{1}{2}$$$ und $$$f{\left(x \right)} = 4 \cos{\left(x \right)} + 4 \cos{\left(5 x \right)}$$$ an:

$$\frac{3 \sin{\left(x \right)}}{4} - \frac{\sin{\left(3 x \right)}}{12} - \frac{{\color{red}{\int{\left(2 \cos{\left(x \right)} + 2 \cos{\left(5 x \right)}\right)d x}}}}{8} = \frac{3 \sin{\left(x \right)}}{4} - \frac{\sin{\left(3 x \right)}}{12} - \frac{{\color{red}{\left(\frac{\int{\left(4 \cos{\left(x \right)} + 4 \cos{\left(5 x \right)}\right)d x}}{2}\right)}}}{8}$$

Gliedweise integrieren:

$$\frac{3 \sin{\left(x \right)}}{4} - \frac{\sin{\left(3 x \right)}}{12} - \frac{{\color{red}{\int{\left(4 \cos{\left(x \right)} + 4 \cos{\left(5 x \right)}\right)d x}}}}{16} = \frac{3 \sin{\left(x \right)}}{4} - \frac{\sin{\left(3 x \right)}}{12} - \frac{{\color{red}{\left(\int{4 \cos{\left(x \right)} d x} + \int{4 \cos{\left(5 x \right)} d x}\right)}}}{16}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=4$$$ und $$$f{\left(x \right)} = \cos{\left(x \right)}$$$ an:

$$\frac{3 \sin{\left(x \right)}}{4} - \frac{\sin{\left(3 x \right)}}{12} - \frac{\int{4 \cos{\left(5 x \right)} d x}}{16} - \frac{{\color{red}{\int{4 \cos{\left(x \right)} d x}}}}{16} = \frac{3 \sin{\left(x \right)}}{4} - \frac{\sin{\left(3 x \right)}}{12} - \frac{\int{4 \cos{\left(5 x \right)} d x}}{16} - \frac{{\color{red}{\left(4 \int{\cos{\left(x \right)} d x}\right)}}}{16}$$

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

$$\frac{3 \sin{\left(x \right)}}{4} - \frac{\sin{\left(3 x \right)}}{12} - \frac{\int{4 \cos{\left(5 x \right)} d x}}{16} - \frac{{\color{red}{\int{\cos{\left(x \right)} d x}}}}{4} = \frac{3 \sin{\left(x \right)}}{4} - \frac{\sin{\left(3 x \right)}}{12} - \frac{\int{4 \cos{\left(5 x \right)} d x}}{16} - \frac{{\color{red}{\sin{\left(x \right)}}}}{4}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=4$$$ und $$$f{\left(x \right)} = \cos{\left(5 x \right)}$$$ an:

$$\frac{\sin{\left(x \right)}}{2} - \frac{\sin{\left(3 x \right)}}{12} - \frac{{\color{red}{\int{4 \cos{\left(5 x \right)} d x}}}}{16} = \frac{\sin{\left(x \right)}}{2} - \frac{\sin{\left(3 x \right)}}{12} - \frac{{\color{red}{\left(4 \int{\cos{\left(5 x \right)} d x}\right)}}}{16}$$

Sei $$$v=5 x$$$.

Dann $$$dv=\left(5 x\right)^{\prime }dx = 5 dx$$$ (die Schritte sind » zu sehen), und es gilt $$$dx = \frac{dv}{5}$$$.

Das Integral lässt sich umschreiben als

$$\frac{\sin{\left(x \right)}}{2} - \frac{\sin{\left(3 x \right)}}{12} - \frac{{\color{red}{\int{\cos{\left(5 x \right)} d x}}}}{4} = \frac{\sin{\left(x \right)}}{2} - \frac{\sin{\left(3 x \right)}}{12} - \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{5} d v}}}}{4}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ mit $$$c=\frac{1}{5}$$$ und $$$f{\left(v \right)} = \cos{\left(v \right)}$$$ an:

$$\frac{\sin{\left(x \right)}}{2} - \frac{\sin{\left(3 x \right)}}{12} - \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{5} d v}}}}{4} = \frac{\sin{\left(x \right)}}{2} - \frac{\sin{\left(3 x \right)}}{12} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{5}\right)}}}{4}$$

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

$$\frac{\sin{\left(x \right)}}{2} - \frac{\sin{\left(3 x \right)}}{12} - \frac{{\color{red}{\int{\cos{\left(v \right)} d v}}}}{20} = \frac{\sin{\left(x \right)}}{2} - \frac{\sin{\left(3 x \right)}}{12} - \frac{{\color{red}{\sin{\left(v \right)}}}}{20}$$

Zur Erinnerung: $$$v=5 x$$$:

$$\frac{\sin{\left(x \right)}}{2} - \frac{\sin{\left(3 x \right)}}{12} - \frac{\sin{\left({\color{red}{v}} \right)}}{20} = \frac{\sin{\left(x \right)}}{2} - \frac{\sin{\left(3 x \right)}}{12} - \frac{\sin{\left({\color{red}{\left(5 x\right)}} \right)}}{20}$$

Daher,

$$\int{4 \sin^{2}{\left(x \right)} \cos^{3}{\left(x \right)} d x} = \frac{\sin{\left(x \right)}}{2} - \frac{\sin{\left(3 x \right)}}{12} - \frac{\sin{\left(5 x \right)}}{20}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{4 \sin^{2}{\left(x \right)} \cos^{3}{\left(x \right)} d x} = \frac{\sin{\left(x \right)}}{2} - \frac{\sin{\left(3 x \right)}}{12} - \frac{\sin{\left(5 x \right)}}{20}+C$$

$$$\int 4 \sin^{2}{\left(x \right)} \cos^{3}{\left(x \right)}\, dx = \left(\frac{\sin{\left(x \right)}}{2} - \frac{\sin{\left(3 x \right)}}{12} - \frac{\sin{\left(5 x \right)}}{20}\right) + C$$$A