Integral von $$$8 \sin^{3}{\left(x \right)}$$$

Bestimme $$$\int 8 \sin^{3}{\left(x \right)}\, dx$$$.

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=8$$$ und $$$f{\left(x \right)} = \sin^{3}{\left(x \right)}$$$ an:

$${\color{red}{\int{8 \sin^{3}{\left(x \right)} d x}}} = {\color{red}{\left(8 \int{\sin^{3}{\left(x \right)} d x}\right)}}$$

Klammern Sie einen Sinus aus und drücken Sie alles andere durch den Kosinus aus, unter Verwendung der Formel $$$\sin^2\left(\alpha \right)=-\cos^2\left(\alpha \right)+1$$$ mit $$$\alpha=x$$$:

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

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

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

Somit,

$$8 {\color{red}{\int{\left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} d x}}} = 8 {\color{red}{\int{\left(u^{2} - 1\right)d u}}}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=-1$$$ und $$$f{\left(u \right)} = 1 - u^{2}$$$ an:

$$8 {\color{red}{\int{\left(u^{2} - 1\right)d u}}} = 8 {\color{red}{\left(- \int{\left(1 - u^{2}\right)d u}\right)}}$$

Gliedweise integrieren:

$$- 8 {\color{red}{\int{\left(1 - u^{2}\right)d u}}} = - 8 {\color{red}{\left(\int{1 d u} - \int{u^{2} d u}\right)}}$$

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

$$8 \int{u^{2} d u} - 8 {\color{red}{\int{1 d u}}} = 8 \int{u^{2} d u} - 8 {\color{red}{u}}$$

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

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

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

$$- 8 {\color{red}{u}} + \frac{8 {\color{red}{u}}^{3}}{3} = - 8 {\color{red}{\cos{\left(x \right)}}} + \frac{8 {\color{red}{\cos{\left(x \right)}}}^{3}}{3}$$

Daher,

$$\int{8 \sin^{3}{\left(x \right)} d x} = \frac{8 \cos^{3}{\left(x \right)}}{3} - 8 \cos{\left(x \right)}$$

Vereinfachen:

$$\int{8 \sin^{3}{\left(x \right)} d x} = \frac{8 \left(\cos^{2}{\left(x \right)} - 3\right) \cos{\left(x \right)}}{3}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{8 \sin^{3}{\left(x \right)} d x} = \frac{8 \left(\cos^{2}{\left(x \right)} - 3\right) \cos{\left(x \right)}}{3}+C$$

$$$\int 8 \sin^{3}{\left(x \right)}\, dx = \frac{8 \left(\cos^{2}{\left(x \right)} - 3\right) \cos{\left(x \right)}}{3} + C$$$A