COMEDK · Maths · 28. Indefinite Integration
If \(x=2 a \sin ^{-1}\left(\sqrt{\dfrac{y}{2 a}}\right)-\sqrt{2 a y-y^2}\) then \(\dfrac{d y}{d x}=\)
- A \(\sqrt{\dfrac{y}{2 a-y}}\)
- B \(\dfrac{\sqrt{2 a-y}}{y}\)
- C \(\dfrac{1}{2}\left(\sqrt{\dfrac{2 a-y}{y}}\right)\)
- D \(\sqrt{\dfrac{2 a-y}{y}}\)
Answer & Solution
Correct Answer
(D) \(\sqrt{\dfrac{2 a-y}{y}}\)
Step-by-step Solution
Detailed explanation
Given \(x = 2a \sin^{-1}\left(\sqrt{\dfrac{y}{2a}}\right) - \sqrt{2ay - y^2}\). Let \(y = 2a \sin^2 \theta\). Then \(\sqrt{\dfrac{y}{2a}} = \sin \theta\), so \(\theta = \sin^{-1}\left(\sqrt{\dfrac{y}{2a}}\right)\). Also,…
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