Integral von $$$\frac{1}{\sqrt{- x^{2} + x}}$$$

Bestimme $$$\int \frac{1}{\sqrt{- x^{2} + x}}\, dx$$$.

Quadrat ergänzen (Schritte siehe »): $$$- x^{2} + x = \frac{1}{4} - \left(x - \frac{1}{2}\right)^{2}$$$:

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

Sei $$$u=x - \frac{1}{2}$$$.

Dann $$$du=\left(x - \frac{1}{2}\right)^{\prime }dx = 1 dx$$$ (die Schritte sind » zu sehen), und es gilt $$$dx = du$$$.

Also,

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

Sei $$$u=\frac{\sin{\left(v \right)}}{2}$$$.

Dann $$$du=\left(\frac{\sin{\left(v \right)}}{2}\right)^{\prime }dv = \frac{\cos{\left(v \right)}}{2} dv$$$ (die Schritte sind » zu sehen).

Somit folgt, dass $$$v=\operatorname{asin}{\left(2 u \right)}$$$.

Also,

$$$\frac{1}{\sqrt{\frac{1}{4} - u ^{2}}} = \frac{1}{\sqrt{\frac{1}{4} - \frac{\sin^{2}{\left( v \right)}}{4}}}$$$

Verwenden Sie die Identität $$$1 - \sin^{2}{\left( v \right)} = \cos^{2}{\left( v \right)}$$$:

$$$\frac{1}{\sqrt{\frac{1}{4} - \frac{\sin^{2}{\left( v \right)}}{4}}}=\frac{2}{\sqrt{1 - \sin^{2}{\left( v \right)}}}=\frac{2}{\sqrt{\cos^{2}{\left( v \right)}}}$$$

Setzen wir $$$\cos{\left( v \right)} \ge 0$$$ voraus, so erhalten wir Folgendes:

$$$\frac{2}{\sqrt{\cos^{2}{\left( v \right)}}} = \frac{2}{\cos{\left( v \right)}}$$$

Das Integral wird zu

$${\color{red}{\int{\frac{1}{\sqrt{\frac{1}{4} - u^{2}}} d u}}} = {\color{red}{\int{1 d v}}}$$

Wenden Sie die Konstantenregel $$$\int c\, dv = c v$$$ mit $$$c=1$$$ an:

$${\color{red}{\int{1 d v}}} = {\color{red}{v}}$$

Zur Erinnerung: $$$v=\operatorname{asin}{\left(2 u \right)}$$$:

$${\color{red}{v}} = {\color{red}{\operatorname{asin}{\left(2 u \right)}}}$$

Zur Erinnerung: $$$u=x - \frac{1}{2}$$$:

$$\operatorname{asin}{\left(2 {\color{red}{u}} \right)} = \operatorname{asin}{\left(2 {\color{red}{\left(x - \frac{1}{2}\right)}} \right)}$$

Daher,

$$\int{\frac{1}{\sqrt{- x^{2} + x}} d x} = \operatorname{asin}{\left(2 x - 1 \right)}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{1}{\sqrt{- x^{2} + x}} d x} = \operatorname{asin}{\left(2 x - 1 \right)}+C$$

$$$\int \frac{1}{\sqrt{- x^{2} + x}}\, dx = \operatorname{asin}{\left(2 x - 1 \right)} + C$$$A