Question

$\int 4 \cos \left(x + \frac{\pi}{6}\right) \cos 2x . \cos\left(\frac{5\pi}{6} + x\right)dx $

• A. $-\left(x - \frac{\sin4x}{4} - \frac{\sin2x}{2}\right) +c $
• B. $-\left(x + \frac{\sin4x}{4} - \frac{\sin2x}{2}\right) +c $
• C. $-\left(x - \frac{\sin4x}{4} + \frac{\sin2x}{2}\right) +c $
• D. $-\left(x + \frac{\sin4x}{4} + \frac{\sin2x}{2}\right) +c $

Answer 1

Answer: (A)

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