Integral von $$$\sqrt[4]{\sin{\left(x \right)}} \cos^{3}{\left(x \right)}$$$

Bestimme $$$\int \sqrt[4]{\sin{\left(x \right)}} \cos^{3}{\left(x \right)}\, dx$$$.

Klammern Sie einen Kosinus aus und drücken Sie alles Übrige in Abhängigkeit vom Sinus aus, mithilfe der Formel $$$\cos^2\left(\alpha \right)=-\sin^2\left(\alpha \right)+1$$$ mit $$$\alpha=x$$$.:

$${\color{red}{\int{\sqrt[4]{\sin{\left(x \right)}} \cos^{3}{\left(x \right)} d x}}} = {\color{red}{\int{\left(1 - \sin^{2}{\left(x \right)}\right) \sqrt[4]{\sin{\left(x \right)}} \cos{\left(x \right)} d x}}}$$

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

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

Das Integral wird zu

$${\color{red}{\int{\left(1 - \sin^{2}{\left(x \right)}\right) \sqrt[4]{\sin{\left(x \right)}} \cos{\left(x \right)} d x}}} = {\color{red}{\int{\sqrt[4]{u} \left(1 - u^{2}\right) d u}}}$$

Expand the expression:

$${\color{red}{\int{\sqrt[4]{u} \left(1 - u^{2}\right) d u}}} = {\color{red}{\int{\left(- u^{\frac{9}{4}} + \sqrt[4]{u}\right)d u}}}$$

Gliedweise integrieren:

$${\color{red}{\int{\left(- u^{\frac{9}{4}} + \sqrt[4]{u}\right)d u}}} = {\color{red}{\left(\int{\sqrt[4]{u} d u} - \int{u^{\frac{9}{4}} d u}\right)}}$$

Wenden Sie die Potenzregel $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=\frac{1}{4}$$$ an:

$$- \int{u^{\frac{9}{4}} d u} + {\color{red}{\int{\sqrt[4]{u} d u}}}=- \int{u^{\frac{9}{4}} d u} + {\color{red}{\int{u^{\frac{1}{4}} d u}}}=- \int{u^{\frac{9}{4}} d u} + {\color{red}{\frac{u^{\frac{1}{4} + 1}}{\frac{1}{4} + 1}}}=- \int{u^{\frac{9}{4}} d u} + {\color{red}{\left(\frac{4 u^{\frac{5}{4}}}{5}\right)}}$$

Wenden Sie die Potenzregel $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=\frac{9}{4}$$$ an:

$$\frac{4 u^{\frac{5}{4}}}{5} - {\color{red}{\int{u^{\frac{9}{4}} d u}}}=\frac{4 u^{\frac{5}{4}}}{5} - {\color{red}{\frac{u^{1 + \frac{9}{4}}}{1 + \frac{9}{4}}}}=\frac{4 u^{\frac{5}{4}}}{5} - {\color{red}{\left(\frac{4 u^{\frac{13}{4}}}{13}\right)}}$$

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

$$\frac{4 {\color{red}{u}}^{\frac{5}{4}}}{5} - \frac{4 {\color{red}{u}}^{\frac{13}{4}}}{13} = \frac{4 {\color{red}{\sin{\left(x \right)}}}^{\frac{5}{4}}}{5} - \frac{4 {\color{red}{\sin{\left(x \right)}}}^{\frac{13}{4}}}{13}$$

Daher,

$$\int{\sqrt[4]{\sin{\left(x \right)}} \cos^{3}{\left(x \right)} d x} = - \frac{4 \sin^{\frac{13}{4}}{\left(x \right)}}{13} + \frac{4 \sin^{\frac{5}{4}}{\left(x \right)}}{5}$$

Vereinfachen:

$$\int{\sqrt[4]{\sin{\left(x \right)}} \cos^{3}{\left(x \right)} d x} = \frac{4 \left(13 - 5 \sin^{2}{\left(x \right)}\right) \sin^{\frac{5}{4}}{\left(x \right)}}{65}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\sqrt[4]{\sin{\left(x \right)}} \cos^{3}{\left(x \right)} d x} = \frac{4 \left(13 - 5 \sin^{2}{\left(x \right)}\right) \sin^{\frac{5}{4}}{\left(x \right)}}{65}+C$$

$$$\int \sqrt[4]{\sin{\left(x \right)}} \cos^{3}{\left(x \right)}\, dx = \frac{4 \left(13 - 5 \sin^{2}{\left(x \right)}\right) \sin^{\frac{5}{4}}{\left(x \right)}}{65} + C$$$A