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

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

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 $$$\alpha=x$$$:

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

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,

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

Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=-1$$$ und $$$f{\left(u \right)} = u^{3} \left(1 - u^{2}\right)$$$ an:

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

Expand the expression:

$$- {\color{red}{\int{u^{3} \left(1 - u^{2}\right) d u}}} = - {\color{red}{\int{\left(- u^{5} + u^{3}\right)d u}}}$$

Gliedweise integrieren:

$$- {\color{red}{\int{\left(- u^{5} + u^{3}\right)d u}}} = - {\color{red}{\left(\int{u^{3} d u} - \int{u^{5} d u}\right)}}$$

Wenden Sie die Potenzregel $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=3$$$ an:

$$\int{u^{5} d u} - {\color{red}{\int{u^{3} d u}}}=\int{u^{5} d u} - {\color{red}{\frac{u^{1 + 3}}{1 + 3}}}=\int{u^{5} d u} - {\color{red}{\left(\frac{u^{4}}{4}\right)}}$$

Wenden Sie die Potenzregel $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=5$$$ an:

$$- \frac{u^{4}}{4} + {\color{red}{\int{u^{5} d u}}}=- \frac{u^{4}}{4} + {\color{red}{\frac{u^{1 + 5}}{1 + 5}}}=- \frac{u^{4}}{4} + {\color{red}{\left(\frac{u^{6}}{6}\right)}}$$

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

$$- \frac{{\color{red}{u}}^{4}}{4} + \frac{{\color{red}{u}}^{6}}{6} = - \frac{{\color{red}{\cos{\left(x \right)}}}^{4}}{4} + \frac{{\color{red}{\cos{\left(x \right)}}}^{6}}{6}$$

Daher,

$$\int{\sin^{3}{\left(x \right)} \cos^{3}{\left(x \right)} d x} = \frac{\cos^{6}{\left(x \right)}}{6} - \frac{\cos^{4}{\left(x \right)}}{4}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\sin^{3}{\left(x \right)} \cos^{3}{\left(x \right)} d x} = \frac{\cos^{6}{\left(x \right)}}{6} - \frac{\cos^{4}{\left(x \right)}}{4}+C$$

$$$\int \sin^{3}{\left(x \right)} \cos^{3}{\left(x \right)}\, dx = \left(\frac{\cos^{6}{\left(x \right)}}{6} - \frac{\cos^{4}{\left(x \right)}}{4}\right) + C$$$A