Integral von $$$\left(2 - 3 \sin{\left(x \right)}\right) \cos^{2}{\left(x \right)}$$$
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Ihre Eingabe
Bestimme $$$\int \left(2 - 3 \sin{\left(x \right)}\right) \cos^{2}{\left(x \right)}\, dx$$$.
Lösung
Wende die Potenzreduktionsformel $$$\cos^{2}{\left(\alpha \right)} = \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{1}{2}$$$ mit $$$\alpha=x$$$ an:
$${\color{red}{\int{\left(2 - 3 \sin{\left(x \right)}\right) \cos^{2}{\left(x \right)} d x}}} = {\color{red}{\int{\frac{\left(2 - 3 \sin{\left(x \right)}\right) \left(\cos{\left(2 x \right)} + 1\right)}{2} d x}}}$$
Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=\frac{1}{2}$$$ und $$$f{\left(x \right)} = \left(2 - 3 \sin{\left(x \right)}\right) \left(\cos{\left(2 x \right)} + 1\right)$$$ an:
$${\color{red}{\int{\frac{\left(2 - 3 \sin{\left(x \right)}\right) \left(\cos{\left(2 x \right)} + 1\right)}{2} d x}}} = {\color{red}{\left(\frac{\int{\left(2 - 3 \sin{\left(x \right)}\right) \left(\cos{\left(2 x \right)} + 1\right) d x}}{2}\right)}}$$
Expand the expression:
$$\frac{{\color{red}{\int{\left(2 - 3 \sin{\left(x \right)}\right) \left(\cos{\left(2 x \right)} + 1\right) d x}}}}{2} = \frac{{\color{red}{\int{\left(- 3 \sin{\left(x \right)} \cos{\left(2 x \right)} - 3 \sin{\left(x \right)} + 2 \cos{\left(2 x \right)} + 2\right)d x}}}}{2}$$
Gliedweise integrieren:
$$\frac{{\color{red}{\int{\left(- 3 \sin{\left(x \right)} \cos{\left(2 x \right)} - 3 \sin{\left(x \right)} + 2 \cos{\left(2 x \right)} + 2\right)d x}}}}{2} = \frac{{\color{red}{\left(\int{2 d x} - \int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x} - \int{3 \sin{\left(x \right)} d x} + \int{2 \cos{\left(2 x \right)} d x}\right)}}}{2}$$
Wenden Sie die Konstantenregel $$$\int c\, dx = c x$$$ mit $$$c=2$$$ an:
$$- \frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{2} - \frac{\int{3 \sin{\left(x \right)} d x}}{2} + \frac{\int{2 \cos{\left(2 x \right)} d x}}{2} + \frac{{\color{red}{\int{2 d x}}}}{2} = - \frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{2} - \frac{\int{3 \sin{\left(x \right)} d x}}{2} + \frac{\int{2 \cos{\left(2 x \right)} d x}}{2} + \frac{{\color{red}{\left(2 x\right)}}}{2}$$
Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=3$$$ und $$$f{\left(x \right)} = \sin{\left(x \right)}$$$ an:
$$x - \frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{2} + \frac{\int{2 \cos{\left(2 x \right)} d x}}{2} - \frac{{\color{red}{\int{3 \sin{\left(x \right)} d x}}}}{2} = x - \frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{2} + \frac{\int{2 \cos{\left(2 x \right)} d x}}{2} - \frac{{\color{red}{\left(3 \int{\sin{\left(x \right)} d x}\right)}}}{2}$$
Das Integral des Sinus lautet $$$\int{\sin{\left(x \right)} d x} = - \cos{\left(x \right)}$$$:
$$x - \frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{2} + \frac{\int{2 \cos{\left(2 x \right)} d x}}{2} - \frac{3 {\color{red}{\int{\sin{\left(x \right)} d x}}}}{2} = x - \frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{2} + \frac{\int{2 \cos{\left(2 x \right)} d x}}{2} - \frac{3 {\color{red}{\left(- \cos{\left(x \right)}\right)}}}{2}$$
Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=2$$$ und $$$f{\left(x \right)} = \cos{\left(2 x \right)}$$$ an:
$$x + \frac{3 \cos{\left(x \right)}}{2} - \frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{2} + \frac{{\color{red}{\int{2 \cos{\left(2 x \right)} d x}}}}{2} = x + \frac{3 \cos{\left(x \right)}}{2} - \frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{2} + \frac{{\color{red}{\left(2 \int{\cos{\left(2 x \right)} d x}\right)}}}{2}$$
Sei $$$u=2 x$$$.
Dann $$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (die Schritte sind » zu sehen), und es gilt $$$dx = \frac{du}{2}$$$.
Also,
$$x + \frac{3 \cos{\left(x \right)}}{2} - \frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{2} + {\color{red}{\int{\cos{\left(2 x \right)} d x}}} = x + \frac{3 \cos{\left(x \right)}}{2} - \frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{2} + {\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}$$
Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=\frac{1}{2}$$$ und $$$f{\left(u \right)} = \cos{\left(u \right)}$$$ an:
$$x + \frac{3 \cos{\left(x \right)}}{2} - \frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{2} + {\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}} = x + \frac{3 \cos{\left(x \right)}}{2} - \frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{2} + {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{2}\right)}}$$
Das Integral des Kosinus ist $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:
$$x + \frac{3 \cos{\left(x \right)}}{2} - \frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{2} + \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{2} = x + \frac{3 \cos{\left(x \right)}}{2} - \frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{2} + \frac{{\color{red}{\sin{\left(u \right)}}}}{2}$$
Zur Erinnerung: $$$u=2 x$$$:
$$x + \frac{3 \cos{\left(x \right)}}{2} - \frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{2} + \frac{\sin{\left({\color{red}{u}} \right)}}{2} = x + \frac{3 \cos{\left(x \right)}}{2} - \frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{2} + \frac{\sin{\left({\color{red}{\left(2 x\right)}} \right)}}{2}$$
Schreibe $$$\sin\left(x \right)\cos\left(2 x \right)$$$ mithilfe der Formel $$$\sin\left(\alpha \right)\cos\left(\beta \right)=\frac{1}{2} \sin\left(\alpha-\beta \right)+\frac{1}{2} \sin\left(\alpha+\beta \right)$$$ mit $$$\alpha=x$$$ und $$$\beta=2 x$$$ um:
$$x + \frac{\sin{\left(2 x \right)}}{2} + \frac{3 \cos{\left(x \right)}}{2} - \frac{{\color{red}{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}}}{2} = x + \frac{\sin{\left(2 x \right)}}{2} + \frac{3 \cos{\left(x \right)}}{2} - \frac{{\color{red}{\int{\left(- \frac{3 \sin{\left(x \right)}}{2} + \frac{3 \sin{\left(3 x \right)}}{2}\right)d x}}}}{2}$$
Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=\frac{1}{2}$$$ und $$$f{\left(x \right)} = - 3 \sin{\left(x \right)} + 3 \sin{\left(3 x \right)}$$$ an:
$$x + \frac{\sin{\left(2 x \right)}}{2} + \frac{3 \cos{\left(x \right)}}{2} - \frac{{\color{red}{\int{\left(- \frac{3 \sin{\left(x \right)}}{2} + \frac{3 \sin{\left(3 x \right)}}{2}\right)d x}}}}{2} = x + \frac{\sin{\left(2 x \right)}}{2} + \frac{3 \cos{\left(x \right)}}{2} - \frac{{\color{red}{\left(\frac{\int{\left(- 3 \sin{\left(x \right)} + 3 \sin{\left(3 x \right)}\right)d x}}{2}\right)}}}{2}$$
Gliedweise integrieren:
$$x + \frac{\sin{\left(2 x \right)}}{2} + \frac{3 \cos{\left(x \right)}}{2} - \frac{{\color{red}{\int{\left(- 3 \sin{\left(x \right)} + 3 \sin{\left(3 x \right)}\right)d x}}}}{4} = x + \frac{\sin{\left(2 x \right)}}{2} + \frac{3 \cos{\left(x \right)}}{2} - \frac{{\color{red}{\left(- \int{3 \sin{\left(x \right)} d x} + \int{3 \sin{\left(3 x \right)} d x}\right)}}}{4}$$
Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=3$$$ und $$$f{\left(x \right)} = \sin{\left(x \right)}$$$ an:
$$x + \frac{\sin{\left(2 x \right)}}{2} + \frac{3 \cos{\left(x \right)}}{2} - \frac{\int{3 \sin{\left(3 x \right)} d x}}{4} + \frac{{\color{red}{\int{3 \sin{\left(x \right)} d x}}}}{4} = x + \frac{\sin{\left(2 x \right)}}{2} + \frac{3 \cos{\left(x \right)}}{2} - \frac{\int{3 \sin{\left(3 x \right)} d x}}{4} + \frac{{\color{red}{\left(3 \int{\sin{\left(x \right)} d x}\right)}}}{4}$$
Das Integral des Sinus lautet $$$\int{\sin{\left(x \right)} d x} = - \cos{\left(x \right)}$$$:
$$x + \frac{\sin{\left(2 x \right)}}{2} + \frac{3 \cos{\left(x \right)}}{2} - \frac{\int{3 \sin{\left(3 x \right)} d x}}{4} + \frac{3 {\color{red}{\int{\sin{\left(x \right)} d x}}}}{4} = x + \frac{\sin{\left(2 x \right)}}{2} + \frac{3 \cos{\left(x \right)}}{2} - \frac{\int{3 \sin{\left(3 x \right)} d x}}{4} + \frac{3 {\color{red}{\left(- \cos{\left(x \right)}\right)}}}{4}$$
Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=3$$$ und $$$f{\left(x \right)} = \sin{\left(3 x \right)}$$$ an:
$$x + \frac{\sin{\left(2 x \right)}}{2} + \frac{3 \cos{\left(x \right)}}{4} - \frac{{\color{red}{\int{3 \sin{\left(3 x \right)} d x}}}}{4} = x + \frac{\sin{\left(2 x \right)}}{2} + \frac{3 \cos{\left(x \right)}}{4} - \frac{{\color{red}{\left(3 \int{\sin{\left(3 x \right)} d x}\right)}}}{4}$$
Sei $$$u=3 x$$$.
Dann $$$du=\left(3 x\right)^{\prime }dx = 3 dx$$$ (die Schritte sind » zu sehen), und es gilt $$$dx = \frac{du}{3}$$$.
Also,
$$x + \frac{\sin{\left(2 x \right)}}{2} + \frac{3 \cos{\left(x \right)}}{4} - \frac{3 {\color{red}{\int{\sin{\left(3 x \right)} d x}}}}{4} = x + \frac{\sin{\left(2 x \right)}}{2} + \frac{3 \cos{\left(x \right)}}{4} - \frac{3 {\color{red}{\int{\frac{\sin{\left(u \right)}}{3} d u}}}}{4}$$
Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=\frac{1}{3}$$$ und $$$f{\left(u \right)} = \sin{\left(u \right)}$$$ an:
$$x + \frac{\sin{\left(2 x \right)}}{2} + \frac{3 \cos{\left(x \right)}}{4} - \frac{3 {\color{red}{\int{\frac{\sin{\left(u \right)}}{3} d u}}}}{4} = x + \frac{\sin{\left(2 x \right)}}{2} + \frac{3 \cos{\left(x \right)}}{4} - \frac{3 {\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{3}\right)}}}{4}$$
Das Integral des Sinus lautet $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$:
$$x + \frac{\sin{\left(2 x \right)}}{2} + \frac{3 \cos{\left(x \right)}}{4} - \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{4} = x + \frac{\sin{\left(2 x \right)}}{2} + \frac{3 \cos{\left(x \right)}}{4} - \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{4}$$
Zur Erinnerung: $$$u=3 x$$$:
$$x + \frac{\sin{\left(2 x \right)}}{2} + \frac{3 \cos{\left(x \right)}}{4} + \frac{\cos{\left({\color{red}{u}} \right)}}{4} = x + \frac{\sin{\left(2 x \right)}}{2} + \frac{3 \cos{\left(x \right)}}{4} + \frac{\cos{\left({\color{red}{\left(3 x\right)}} \right)}}{4}$$
Daher,
$$\int{\left(2 - 3 \sin{\left(x \right)}\right) \cos^{2}{\left(x \right)} d x} = x + \frac{\sin{\left(2 x \right)}}{2} + \frac{3 \cos{\left(x \right)}}{4} + \frac{\cos{\left(3 x \right)}}{4}$$
Fügen Sie die Integrationskonstante hinzu:
$$\int{\left(2 - 3 \sin{\left(x \right)}\right) \cos^{2}{\left(x \right)} d x} = x + \frac{\sin{\left(2 x \right)}}{2} + \frac{3 \cos{\left(x \right)}}{4} + \frac{\cos{\left(3 x \right)}}{4}+C$$
Antwort
$$$\int \left(2 - 3 \sin{\left(x \right)}\right) \cos^{2}{\left(x \right)}\, dx = \left(x + \frac{\sin{\left(2 x \right)}}{2} + \frac{3 \cos{\left(x \right)}}{4} + \frac{\cos{\left(3 x \right)}}{4}\right) + C$$$A