Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

De onbepaalde integraal $\int \frac{4}{\sqrt{4-x^2}}\: dx$ is gelijk aan	A. Bgsin + C
	B. 2.Bgsin + C
	C. 2.Bgsin x + C
	D. 4.Bgsin + C
	E. $\boldsymbol {4 \cdot \ln \| x + \sqrt {4-x^2} \| + C} $

[ 6-3436 - op net sinds 16.3.02-(E)-16.11.2024 ]

Translation in E N G L I S H

The indefinite integral $\int \frac{4}{\sqrt{4-x^2}}\: dx$ equals	A. Arcsin + C
	B. 2.Arcsin + C
	C. 2.Arcsin x + C
	D. 4.Arcsin + C
	E. $\small \boldsymbol {4 \cdot \ln \| x + \sqrt {4-x^2} \| + C} $

Oplossing - Solution

$\begin{aligned} & \text{We steunen op de basisintegraal}\int \frac{1}{\sqrt{1-f^2}}\: df=Bgsin\:f + C\\ & \int \frac{4}{\sqrt{4-x^2}}\: dx =\int \frac{4}{\sqrt{4\left (1-(\frac{x}{2})^2 \right )}}\: dx= \int \frac{4}{2\sqrt{1-(\frac{x}{2})^2 }}\: dx\\ & =2\int \frac{1}{\sqrt{1-(\frac{x}{2})^2 }}\: dx= 4\int \frac{1}{\sqrt{1-(\frac{x}{2})^2 }}\: d(\frac x2)= 4.Bgsin\: \frac x2 + C \end{aligned}$