CUET · MATHS · PYQ PAPER 2025
\(\int \frac{\sin 2 x d x}{\sqrt{9-\cos ^4 x}}\) equals
- A \(\sin ^{-1}\left(\frac{\cos ^2 x}{3}\right)+C: C\) is an arbitrary constant
- B \(\cos ^{-1}\left(\frac{\sin ^2 x}{3}\right)+C: C\) is an arbitrary constant
- C \(\sin \left(\frac{\cos ^2 x}{3}\right)+C: C\) is an arbitrary constant
- D \(-\sin ^{-1}\left(\frac{\cos ^2 x}{3}\right)+C: C\) is an arbitrary constant
Answer & Solution
Correct Answer
(D) \(-\sin ^{-1}\left(\frac{\cos ^2 x}{3}\right)+C: C\) is an arbitrary constant
Step-by-step Solution
Detailed explanation
Let \(u = \cos^2 x\). Then \(du = 2 \cos x (-\sin x) dx = -2 \sin x \cos x dx = -\sin 2x dx\). \(\int \frac{\sin 2 x d x}{\sqrt{9-\cos ^4 x}} = \int \frac{-du}{\sqrt{9-u^2}}\) \(= -\sin^{-1}\left(\frac{u}{3}\right) + C\) \(= -\sin^{-1}\left(\frac{\cos^2 x}{3}\right) + C\)
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