Integral von $$$\sin{\left(t \right)} \sqrt{\cos{\left(t \right)}}$$$

Bestimme $$$\int \sin{\left(t \right)} \sqrt{\cos{\left(t \right)}}\, dt$$$.

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

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

Das Integral lässt sich umschreiben als

$${\color{red}{\int{\sin{\left(t \right)} \sqrt{\cos{\left(t \right)}} d t}}} = {\color{red}{\int{\left(- \sqrt{u}\right)d u}}}$$

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

$${\color{red}{\int{\left(- \sqrt{u}\right)d u}}} = {\color{red}{\left(- \int{\sqrt{u} d u}\right)}}$$

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

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

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

$$- \frac{2 {\color{red}{u}}^{\frac{3}{2}}}{3} = - \frac{2 {\color{red}{\cos{\left(t \right)}}}^{\frac{3}{2}}}{3}$$

Daher,

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

Fügen Sie die Integrationskonstante hinzu:

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

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