CUET · MATHS · PYQ PAPER 2025
\(\int \sin x \sin 2 x \sin 3 x d x\) is equal to:
- A \(-\frac{1}{48}(6 \cos 2 x+3 \cos 4 x+2 \cos 6 x)+C\), Where \(C\) is constant of integration
- B \(-\frac{1}{48}(2 \cos 2 x+3 \cos 4 x-6 \cos 6 x)+C\), Where \(C\) is constant of integration
- C \(-\frac{1}{48}(6 \cos 2 x+3 \cos 4 x-2 \cos 6 x)+C\), Where \(C\) is constant of integration
- D \(-\frac{1}{48}(2 \cos 2 x+3 \cos 4 x+6 \cos 6 x)+C\), Where \(C\) is constant of integration
Answer & Solution
Correct Answer
(C) \(-\frac{1}{48}(6 \cos 2 x+3 \cos 4 x-2 \cos 6 x)+C\), Where \(C\) is constant of integration
Step-by-step Solution
Detailed explanation
\( \int \sin x \sin 2x \sin 3x dx = \int \frac{1}{2}(\cos(x-2x) - \cos(x+2x)) \sin 3x dx \) \( = \frac{1}{2} \int (\cos x - \cos 3x) \sin 3x dx = \frac{1}{2} \int (\sin 3x \cos x - \sin 3x \cos 3x) dx \)…
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