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

# If $\int \mathrm{cos}x\mathrm{cos}2x\mathrm{cos}5xdx$$\begin{array}{r}={A}_{1}\mathrm{sin}2x+{A}_{2}\mathrm{sin}4x+{A}_{3}\mathrm{sin}6x\\ +{A}_{4}\mathrm{sin}8x+C\end{array}$then

1. A

${A}_{1}=\frac{1}{2},{A}_{2}=\frac{1}{4}$

2. B

${A}_{1}=\frac{1}{4},{A}_{2}=\frac{1}{8}$

3. C

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

4. D

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### Solution:

$\begin{array}{r}\left(\mathrm{cos}x\mathrm{cos}2x\right)\mathrm{cos}5x\\ =\frac{1}{2}\left(\mathrm{cos}x+\mathrm{cos}3x\right)\mathrm{cos}5x\end{array}$

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

$\begin{array}{r}\int \mathrm{cos}x\mathrm{cos}2x\mathrm{cos}5xdx=\frac{1}{8}\mathrm{sin}2x+\\ \frac{1}{16}\mathrm{sin}4x+\frac{1}{24}\mathrm{sin}6x+\frac{1}{32}\mathrm{sin}x+C\end{array}$

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