CUET · MATHS · PYQ PAPER 2025
For \(x \in\left(0, \frac{\pi}{2}\right), \int \frac{\sin x+\cos x}{\sqrt{\sin 2 x}} d x\) is equal to
- A \(\sin ^{-1}(\sin 2 x)+C, C\) is an arbitrary constant
- B \(\sin ^{-1}(\cos x-\sin x)+C, C\) is an arbitrary constant
- C \(\sin ^{-1}(\sin x-\cos x)+C, C\) is an arbitrary constant
- D \(\sin ^{-1}(\sin x+\cos x)+C, C\) is an arbitrary constant
Answer & Solution
Correct Answer
(C) \(\sin ^{-1}(\sin x-\cos x)+C, C\) is an arbitrary constant
Step-by-step Solution
Detailed explanation
Let \(t = \sin x - \cos x\). \(dt = (\cos x + \sin x) dx\) \(t^2 = (\sin x - \cos x)^2 = \sin^2 x + \cos^2 x - 2 \sin x \cos x = 1 - \sin 2x\) \(\sin 2x = 1 - t^2\) \(\int \frac{dt}{\sqrt{1 - t^2}}\) \(= \sin^{-1}(t) + C\) \(= \sin^{-1}(\sin x - \cos x) + C\)
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