If ∫ cos x cos 2x cos 5x dx = A<sub>1</sub> sin 2x + A<sub>2</sub> sin 4x + A<sub>3</sub> sin 6x + A<sub>4</sub> sin 8x + c, then the values of A<sub>1</sub>, A<sub>2</sub>, A<sub>3</sub>, A<sub>4</sub> are
\(\cos x\cos 5x = \frac{1}{2}(\cos 4x + \cos 6x)\)
Multiply by \(\cos 2x\): \(\frac{1}{2}(\cos 4x\cos 2x + \cos 6x\cos 2x) = \frac{1}{4}(\cos 2x + \cos 6x + \cos 4x + \cos 8x)\)
Integrate: \(\frac{1}{4}\left(\frac{\sin 2x}{2} + \frac{\sin 4x}{4} + \frac{\sin 6x}{6} + \frac{\sin 8x}{8}\right) + c\)
\(A_1 = \frac{1}{8}\), \(A_2 = \frac{1}{16}\), \(A_3 = \frac{1}{24}\), \(A_4 = \frac{1}{32}\)
Answer: (B)
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