If ∫cos⁡xcos⁡2xcos⁡5xdx=A1sin⁡2x+A2sin⁡4x+A3sin⁡6x+A4sin⁡8x+Cthen

(cos⁡xcos⁡2x)cos⁡5x=12(cos⁡x+cos⁡3x)cos⁡5x=14[cos⁡4x+cos⁡6x]+14[cos⁡2x+cos⁡8x]=14cos⁡2x+14cos⁡4x+14cos⁡6x+14cos⁡8x∫cos⁡xcos⁡2xcos⁡5xdx=18sin⁡2x+116sin⁡4x+124sin⁡6x+132sin⁡x+C

If ∫cos⁡xcos⁡2xcos⁡5xdx=A1sin⁡2x+A2sin⁡4x+A3sin⁡6x+A4sin⁡8x+Cthen

(cos⁡xcos⁡2x)cos⁡5x=12(cos⁡x+cos⁡3x)cos⁡5x=14[cos⁡4x+cos⁡6x]+14[cos⁡2x+cos⁡8x]=14cos⁡2x+14cos⁡4x+14cos⁡6x+14cos⁡8x∫cos⁡xcos⁡2xcos⁡5xdx=18sin⁡2x+116sin⁡4x+124sin⁡6x+132sin⁡x+C

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If $\int \cos x \cos 2 x \cos 5 x d x$
$\begin{aligned} = A_{1} \sin 2 x + A_{2} \sin 4 x + A_{3} \sin 6 x \\ + A_{4} \sin 8 x + C \end{aligned}$
then

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$A_{1} = \frac{1}{4}, A_{2} = \frac{1}{8}$

$A_{2} = \frac{1}{16}, A_{3} = \frac{1}{8}$

$A_{1} = \frac{1}{8}, A_{4} = \frac{1}{32}$

$A_{1} = \frac{1}{2}, A_{2} = \frac{1}{4}$

answer is D.

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Detailed Solution

$\begin{aligned} (\cos x \cos 2 x) \cos 5 x \\ = \frac{1}{2} (\cos x + \cos 3 x) \cos 5 x \end{aligned}$

$\begin{array}{l} = \frac{1}{4} [\cos 4 x + \cos 6 x] + \frac{1}{4} [\cos 2 x + \cos 8 x] \\ = \frac{1}{4} \cos 2 x + \frac{1}{4} \cos 4 x + \frac{1}{4} \cos 6 x \\ + \frac{1}{4} \cos 8 x \end{array}$

$\begin{aligned} \int \cos x \cos 2 x \cos 5 x d x = \frac{1}{8} \sin 2 x + \\ \frac{1}{16} \sin 4 x + \frac{1}{24} \sin 6 x + \frac{1}{32} \sin x + C \end{aligned}$