Integral von $$$\sqrt{2} \left(1 - \sin{\left(2 x \right)}\right)$$$
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Ihre Eingabe
Bestimme $$$\int \sqrt{2} \left(1 - \sin{\left(2 x \right)}\right)\, dx$$$.
Lösung
Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=\sqrt{2}$$$ und $$$f{\left(x \right)} = 1 - \sin{\left(2 x \right)}$$$ an:
$${\color{red}{\int{\sqrt{2} \left(1 - \sin{\left(2 x \right)}\right) d x}}} = {\color{red}{\sqrt{2} \int{\left(1 - \sin{\left(2 x \right)}\right)d x}}}$$
Gliedweise integrieren:
$$\sqrt{2} {\color{red}{\int{\left(1 - \sin{\left(2 x \right)}\right)d x}}} = \sqrt{2} {\color{red}{\left(\int{1 d x} - \int{\sin{\left(2 x \right)} d x}\right)}}$$
Wenden Sie die Konstantenregel $$$\int c\, dx = c x$$$ mit $$$c=1$$$ an:
$$\sqrt{2} \left(- \int{\sin{\left(2 x \right)} d x} + {\color{red}{\int{1 d x}}}\right) = \sqrt{2} \left(- \int{\sin{\left(2 x \right)} d x} + {\color{red}{x}}\right)$$
Sei $$$u=2 x$$$.
Dann $$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (die Schritte sind » zu sehen), und es gilt $$$dx = \frac{du}{2}$$$.
Das Integral wird zu
$$\sqrt{2} \left(x - {\color{red}{\int{\sin{\left(2 x \right)} d x}}}\right) = \sqrt{2} \left(x - {\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}}\right)$$
Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=\frac{1}{2}$$$ und $$$f{\left(u \right)} = \sin{\left(u \right)}$$$ an:
$$\sqrt{2} \left(x - {\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}}\right) = \sqrt{2} \left(x - {\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{2}\right)}}\right)$$
Das Integral des Sinus lautet $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$:
$$\sqrt{2} \left(x - \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{2}\right) = \sqrt{2} \left(x - \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{2}\right)$$
Zur Erinnerung: $$$u=2 x$$$:
$$\sqrt{2} \left(x + \frac{\cos{\left({\color{red}{u}} \right)}}{2}\right) = \sqrt{2} \left(x + \frac{\cos{\left({\color{red}{\left(2 x\right)}} \right)}}{2}\right)$$
Daher,
$$\int{\sqrt{2} \left(1 - \sin{\left(2 x \right)}\right) d x} = \sqrt{2} \left(x + \frac{\cos{\left(2 x \right)}}{2}\right)$$
Fügen Sie die Integrationskonstante hinzu:
$$\int{\sqrt{2} \left(1 - \sin{\left(2 x \right)}\right) d x} = \sqrt{2} \left(x + \frac{\cos{\left(2 x \right)}}{2}\right)+C$$
Antwort
$$$\int \sqrt{2} \left(1 - \sin{\left(2 x \right)}\right)\, dx = \sqrt{2} \left(x + \frac{\cos{\left(2 x \right)}}{2}\right) + C$$$A