Integral von $$$- \sin^{4}{\left(t \right)} + \sin^{2}{\left(t \right)}$$$

Bestimme $$$\int \left(- \sin^{4}{\left(t \right)} + \sin^{2}{\left(t \right)}\right)\, dt$$$.

Gliedweise integrieren:

$${\color{red}{\int{\left(- \sin^{4}{\left(t \right)} + \sin^{2}{\left(t \right)}\right)d t}}} = {\color{red}{\left(\int{\sin^{2}{\left(t \right)} d t} - \int{\sin^{4}{\left(t \right)} d t}\right)}}$$

Wende die Potenzreduktionsformel $$$\sin^{2}{\left(\alpha \right)} = \frac{1}{2} - \frac{\cos{\left(2 \alpha \right)}}{2}$$$ mit $$$\alpha=t$$$ an:

$$- \int{\sin^{4}{\left(t \right)} d t} + {\color{red}{\int{\sin^{2}{\left(t \right)} d t}}} = - \int{\sin^{4}{\left(t \right)} d t} + {\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(2 t \right)}}{2}\right)d t}}}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ mit $$$c=\frac{1}{2}$$$ und $$$f{\left(t \right)} = 1 - \cos{\left(2 t \right)}$$$ an:

$$- \int{\sin^{4}{\left(t \right)} d t} + {\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(2 t \right)}}{2}\right)d t}}} = - \int{\sin^{4}{\left(t \right)} d t} + {\color{red}{\left(\frac{\int{\left(1 - \cos{\left(2 t \right)}\right)d t}}{2}\right)}}$$

Gliedweise integrieren:

$$- \int{\sin^{4}{\left(t \right)} d t} + \frac{{\color{red}{\int{\left(1 - \cos{\left(2 t \right)}\right)d t}}}}{2} = - \int{\sin^{4}{\left(t \right)} d t} + \frac{{\color{red}{\left(\int{1 d t} - \int{\cos{\left(2 t \right)} d t}\right)}}}{2}$$

Wenden Sie die Konstantenregel $$$\int c\, dt = c t$$$ mit $$$c=1$$$ an:

$$- \int{\sin^{4}{\left(t \right)} d t} - \frac{\int{\cos{\left(2 t \right)} d t}}{2} + \frac{{\color{red}{\int{1 d t}}}}{2} = - \int{\sin^{4}{\left(t \right)} d t} - \frac{\int{\cos{\left(2 t \right)} d t}}{2} + \frac{{\color{red}{t}}}{2}$$

Sei $$$u=2 t$$$.

Dann $$$du=\left(2 t\right)^{\prime }dt = 2 dt$$$ (die Schritte sind » zu sehen), und es gilt $$$dt = \frac{du}{2}$$$.

Das Integral lässt sich umschreiben als

$$\frac{t}{2} - \int{\sin^{4}{\left(t \right)} d t} - \frac{{\color{red}{\int{\cos{\left(2 t \right)} d t}}}}{2} = \frac{t}{2} - \int{\sin^{4}{\left(t \right)} d t} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{2}$$

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:

$$\frac{t}{2} - \int{\sin^{4}{\left(t \right)} d t} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{2} = \frac{t}{2} - \int{\sin^{4}{\left(t \right)} d t} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{2}\right)}}}{2}$$

Das Integral des Kosinus ist $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:

$$\frac{t}{2} - \int{\sin^{4}{\left(t \right)} d t} - \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{4} = \frac{t}{2} - \int{\sin^{4}{\left(t \right)} d t} - \frac{{\color{red}{\sin{\left(u \right)}}}}{4}$$

Zur Erinnerung: $$$u=2 t$$$:

$$\frac{t}{2} - \int{\sin^{4}{\left(t \right)} d t} - \frac{\sin{\left({\color{red}{u}} \right)}}{4} = \frac{t}{2} - \int{\sin^{4}{\left(t \right)} d t} - \frac{\sin{\left({\color{red}{\left(2 t\right)}} \right)}}{4}$$

Wende die Potenzreduktionsformel $$$\sin^{4}{\left(\alpha \right)} = - \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{\cos{\left(4 \alpha \right)}}{8} + \frac{3}{8}$$$ mit $$$\alpha=t$$$ an:

$$\frac{t}{2} - \frac{\sin{\left(2 t \right)}}{4} - {\color{red}{\int{\sin^{4}{\left(t \right)} d t}}} = \frac{t}{2} - \frac{\sin{\left(2 t \right)}}{4} - {\color{red}{\int{\left(- \frac{\cos{\left(2 t \right)}}{2} + \frac{\cos{\left(4 t \right)}}{8} + \frac{3}{8}\right)d t}}}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ mit $$$c=\frac{1}{8}$$$ und $$$f{\left(t \right)} = - 4 \cos{\left(2 t \right)} + \cos{\left(4 t \right)} + 3$$$ an:

$$\frac{t}{2} - \frac{\sin{\left(2 t \right)}}{4} - {\color{red}{\int{\left(- \frac{\cos{\left(2 t \right)}}{2} + \frac{\cos{\left(4 t \right)}}{8} + \frac{3}{8}\right)d t}}} = \frac{t}{2} - \frac{\sin{\left(2 t \right)}}{4} - {\color{red}{\left(\frac{\int{\left(- 4 \cos{\left(2 t \right)} + \cos{\left(4 t \right)} + 3\right)d t}}{8}\right)}}$$

Gliedweise integrieren:

$$\frac{t}{2} - \frac{\sin{\left(2 t \right)}}{4} - \frac{{\color{red}{\int{\left(- 4 \cos{\left(2 t \right)} + \cos{\left(4 t \right)} + 3\right)d t}}}}{8} = \frac{t}{2} - \frac{\sin{\left(2 t \right)}}{4} - \frac{{\color{red}{\left(\int{3 d t} - \int{4 \cos{\left(2 t \right)} d t} + \int{\cos{\left(4 t \right)} d t}\right)}}}{8}$$

Wenden Sie die Konstantenregel $$$\int c\, dt = c t$$$ mit $$$c=3$$$ an:

$$\frac{t}{2} - \frac{\sin{\left(2 t \right)}}{4} + \frac{\int{4 \cos{\left(2 t \right)} d t}}{8} - \frac{\int{\cos{\left(4 t \right)} d t}}{8} - \frac{{\color{red}{\int{3 d t}}}}{8} = \frac{t}{2} - \frac{\sin{\left(2 t \right)}}{4} + \frac{\int{4 \cos{\left(2 t \right)} d t}}{8} - \frac{\int{\cos{\left(4 t \right)} d t}}{8} - \frac{{\color{red}{\left(3 t\right)}}}{8}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ mit $$$c=4$$$ und $$$f{\left(t \right)} = \cos{\left(2 t \right)}$$$ an:

$$\frac{t}{8} - \frac{\sin{\left(2 t \right)}}{4} - \frac{\int{\cos{\left(4 t \right)} d t}}{8} + \frac{{\color{red}{\int{4 \cos{\left(2 t \right)} d t}}}}{8} = \frac{t}{8} - \frac{\sin{\left(2 t \right)}}{4} - \frac{\int{\cos{\left(4 t \right)} d t}}{8} + \frac{{\color{red}{\left(4 \int{\cos{\left(2 t \right)} d t}\right)}}}{8}$$

Das Integral $$$\int{\cos{\left(2 t \right)} d t}$$$ wurde bereits berechnet:

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

Daher,

$$\frac{t}{8} - \frac{\sin{\left(2 t \right)}}{4} - \frac{\int{\cos{\left(4 t \right)} d t}}{8} + \frac{{\color{red}{\int{\cos{\left(2 t \right)} d t}}}}{2} = \frac{t}{8} - \frac{\sin{\left(2 t \right)}}{4} - \frac{\int{\cos{\left(4 t \right)} d t}}{8} + \frac{{\color{red}{\left(\frac{\sin{\left(2 t \right)}}{2}\right)}}}{2}$$

Sei $$$v=4 t$$$.

Dann $$$dv=\left(4 t\right)^{\prime }dt = 4 dt$$$ (die Schritte sind » zu sehen), und es gilt $$$dt = \frac{dv}{4}$$$.

Daher,

$$\frac{t}{8} - \frac{{\color{red}{\int{\cos{\left(4 t \right)} d t}}}}{8} = \frac{t}{8} - \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{4} d v}}}}{8}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ mit $$$c=\frac{1}{4}$$$ und $$$f{\left(v \right)} = \cos{\left(v \right)}$$$ an:

$$\frac{t}{8} - \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{4} d v}}}}{8} = \frac{t}{8} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{4}\right)}}}{8}$$

Das Integral des Kosinus ist $$$\int{\cos{\left(v \right)} d v} = \sin{\left(v \right)}$$$:

$$\frac{t}{8} - \frac{{\color{red}{\int{\cos{\left(v \right)} d v}}}}{32} = \frac{t}{8} - \frac{{\color{red}{\sin{\left(v \right)}}}}{32}$$

Zur Erinnerung: $$$v=4 t$$$:

$$\frac{t}{8} - \frac{\sin{\left({\color{red}{v}} \right)}}{32} = \frac{t}{8} - \frac{\sin{\left({\color{red}{\left(4 t\right)}} \right)}}{32}$$

Daher,

$$\int{\left(- \sin^{4}{\left(t \right)} + \sin^{2}{\left(t \right)}\right)d t} = \frac{t}{8} - \frac{\sin{\left(4 t \right)}}{32}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\left(- \sin^{4}{\left(t \right)} + \sin^{2}{\left(t \right)}\right)d t} = \frac{t}{8} - \frac{\sin{\left(4 t \right)}}{32}+C$$

$$$\int \left(- \sin^{4}{\left(t \right)} + \sin^{2}{\left(t \right)}\right)\, dt = \left(\frac{t}{8} - \frac{\sin{\left(4 t \right)}}{32}\right) + C$$$A