Integral von $$$\frac{4}{\sqrt{16 - x^{2}}}$$$

Bestimme $$$\int \frac{4}{\sqrt{16 - x^{2}}}\, dx$$$.

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=4$$$ und $$$f{\left(x \right)} = \frac{1}{\sqrt{16 - x^{2}}}$$$ an:

$${\color{red}{\int{\frac{4}{\sqrt{16 - x^{2}}} d x}}} = {\color{red}{\left(4 \int{\frac{1}{\sqrt{16 - x^{2}}} d x}\right)}}$$

Sei $$$x=4 \sin{\left(u \right)}$$$.

Dann $$$dx=\left(4 \sin{\left(u \right)}\right)^{\prime }du = 4 \cos{\left(u \right)} du$$$ (die Schritte sind » zu sehen).

Somit folgt, dass $$$u=\operatorname{asin}{\left(\frac{x}{4} \right)}$$$.

Somit,

$$$\frac{1}{\sqrt{16 - x^{2}}} = \frac{1}{\sqrt{16 - 16 \sin^{2}{\left( u \right)}}}$$$

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

$$$\frac{1}{\sqrt{16 - 16 \sin^{2}{\left( u \right)}}}=\frac{1}{4 \sqrt{1 - \sin^{2}{\left( u \right)}}}=\frac{1}{4 \sqrt{\cos^{2}{\left( u \right)}}}$$$

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

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

Also,

$$4 {\color{red}{\int{\frac{1}{\sqrt{16 - x^{2}}} d x}}} = 4 {\color{red}{\int{1 d u}}}$$

Wenden Sie die Konstantenregel $$$\int c\, du = c u$$$ mit $$$c=1$$$ an:

$$4 {\color{red}{\int{1 d u}}} = 4 {\color{red}{u}}$$

Zur Erinnerung: $$$u=\operatorname{asin}{\left(\frac{x}{4} \right)}$$$:

$$4 {\color{red}{u}} = 4 {\color{red}{\operatorname{asin}{\left(\frac{x}{4} \right)}}}$$

Daher,

$$\int{\frac{4}{\sqrt{16 - x^{2}}} d x} = 4 \operatorname{asin}{\left(\frac{x}{4} \right)}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{4}{\sqrt{16 - x^{2}}} d x} = 4 \operatorname{asin}{\left(\frac{x}{4} \right)}+C$$

$$$\int \frac{4}{\sqrt{16 - x^{2}}}\, dx = 4 \operatorname{asin}{\left(\frac{x}{4} \right)} + C$$$A