COMEDK · Maths · 28. Indefinite Integration
\(\int \tan^{-1}\left(\sqrt{\dfrac{1 - \sin x}{1 + \sin x}}\right)\, dx =\)
- A \(\dfrac{\pi}{2}x - \dfrac{x^2}{4} + C\)
- B \(\dfrac{\pi}{4} - \dfrac{x}{4} + C\)
- C \(\dfrac{\pi}{4}x - \dfrac{x^2}{2} + C\)
- D \(\dfrac{\pi}{4}x - \dfrac{x^2}{4} + C\)
Answer & Solution
Correct Answer
(D) \(\dfrac{\pi}{4}x - \dfrac{x^2}{4} + C\)
Step-by-step Solution
Detailed explanation
\(\int \tan^{-1}\left(\sqrt{\dfrac{1 - \sin x}{1 + \sin x}}\right)\, dx\) \(\sqrt{\dfrac{1 - \sin x}{1 + \sin x}} = \sqrt{\dfrac{1 - \cos(\pi/2 - x)}{1 + \cos(\pi/2 - x)}}\) Using the half-angle formulas \(1 - \cos \theta = 2 \sin^2(\theta/2)\) and…
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