Q68 / 125 · BITSAT 2010 · Mathematics

\(\int 4 \cos \left(x+\frac{\pi}{6}\right) \cos 2 x \cdot \cos \left(\frac{5 \pi}{6}+x\right) d x\)

  • Correct answer
    A
    \(-\left(x+\frac{\sin 4 x}{4}+\frac{\sin 2 x}{2}\right)+c\)
  • B
    \(-\left(x+\frac{\sin 4 x}{4}-\frac{\sin 2 x}{2}\right)+c\)
  • C
    \(-\left(x-\frac{\sin 4 x}{4}+\frac{\sin 2 x}{2}\right)+c\)
  • D
    \(-\left(x-\frac{\sin 4 x}{4}+\frac{\cos 2 x}{2}\right)+c\)

Solution

\(\begin{aligned} & \int 4 \cos \left(x+\frac{\pi}{6}\right) \cos 2 x \cdot \cos \left(\frac{5 \pi}{6}+x\right) d x \\ & =2 \int\left(\cos (2 x+\pi) \cos \frac{2 \pi}{3}\right) \cos 2 x d x \\ & =2 \int\left(-\cos 2 x-\frac{1}{2}\right) \cos 2 x d x \\ & =\int\left(-2 \cos ^2 2 x-\cos 2 x\right) d x \\ & =-\int(1+\cos 4 x+\cos 2 x) d x \\ & =-x-\frac{\sin 4 x}{x}-\frac{\sin 2 x}{2}+c \end{aligned}\)

Hence, the correct answer is option: 'A'.

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