If int cos x cdot cos 2 x cdot cos 5 x , dx = | vedclass

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(B) Given,$\int \cos x \cdot \cos 2 x \cdot \cos 5 x \, dx$ $= \frac{1}{2} \int (2 \cos 5 x \cos x) \cos 2 x \, dx$ $= \frac{1}{2} \int (\cos 6 x + \c

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$\frac{1}{A} + \frac{1}{D}$

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(B) Given,$\int \cos x \cdot \cos 2 x \cdot \cos 5 x \, dx$ $= \frac{1}{2} \int (2 \cos 5 x \cos x) \cos 2 x \, dx$ $= \frac{1}{2} \int (\cos 6 x + \cos 4 x) \cos 2 x \, dx$ $= \frac{1}{4} \int (2 \cos 6 x \cos 2 x + 2 \cos 4 x \cos 2 x) \, dx$ $= \frac{1}{4} \int (\cos 8 x + \cos 4 x + \cos 6 x + \cos 2 x) \, dx$ $= \frac{1}{4} \left[ \frac{\sin 8 x}{8} + \frac{\sin 4 x}{4} + \frac{\sin 6 x}{6} + \frac{\sin 2 x}{2} \right] + k$ $= \frac{\sin 2 x}{8} + \frac{\sin 4 x}{16} + \frac{\sin 6 x}{24} + \frac{\sin 8 x}{32} + k$ Comparing with $A \sin 2 x + B \sin 4 x + C \sin 6 x + D \sin 8 x + k$,we get: $A = \frac{1}{8}, B = \frac{1}{16}, C = \frac{1}{24}, D = \frac{1}{32}$ Then,$\frac{1}{B} + \frac{1}{C} = 16 + 24 = 40$ And $\frac{1}{A} + \frac{1}{D} = 8 + 32 = 40$ Therefore,$\frac{1}{B} + \frac{1}{C} = \frac{1}{A} + \frac{1}{D}$.

If $\int \cos x \cdot \cos 2 x \cdot \cos 5 x \, dx = A \sin 2 x + B \sin 4 x + C \sin 6 x + D \sin 8 x + k$ (where $k$ is the arbitrary constant of integration),then $\frac{1}{B} + \frac{1}{C} = $

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