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

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

Gliedweise integrieren:

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

Klammern Sie einen Kosinus aus und drücken Sie alles Übrige in Abhängigkeit vom Sinus aus, mithilfe der Formel $$$\cos^2\left(\alpha \right)=-\sin^2\left(\alpha \right)+1$$$ mit $$$\alpha=x$$$.:

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

Sei $$$u=\sin{\left(x \right)}$$$.

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

Das Integral wird zu

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

Gliedweise integrieren:

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

Wenden Sie die Konstantenregel $$$\int c\, du = c u$$$ mit $$$c=1$$$ an:

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

Wenden Sie die Potenzregel $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=2$$$ an:

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

Zur Erinnerung: $$$u=\sin{\left(x \right)}$$$:

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

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

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

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

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

Expand the expression:

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

Gliedweise integrieren:

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

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

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

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

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

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:

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

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

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

Gliedweise integrieren:

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

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

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

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

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

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

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

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,

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

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:

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

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

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

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

$$- \frac{\sin^{3}{\left(x \right)}}{3} + \frac{\sin{\left(x \right)}}{4} + \frac{\sin{\left({\color{red}{u}} \right)}}{4} = - \frac{\sin^{3}{\left(x \right)}}{3} + \frac{\sin{\left(x \right)}}{4} + \frac{\sin{\left({\color{red}{\left(3 x\right)}} \right)}}{4}$$

Daher,

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

Vereinfachen:

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

Fügen Sie die Integrationskonstante hinzu:

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

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