Solveeit Logo
Login

Question

Mathematics Question on Methods of Integration

∫4cos⁡(x+π6)cos⁡2x.cos⁡(5π6+x)dx\int 4 \cos \left(x + \frac{\pi}{6}\right) \cos 2x . \cos\left(\frac{5\pi}{6} + x\right)dx

A

−(x−sin⁡4x4−sin⁡2x2)+c-\left(x - \frac{\sin4x}{4} - \frac{\sin2x}{2}\right) +c

B

−(x+sin⁡4x4−sin⁡2x2)+c-\left(x + \frac{\sin4x}{4} - \frac{\sin2x}{2}\right) +c

C

−(x−sin⁡4x4+sin⁡2x2)+c-\left(x - \frac{\sin4x}{4} + \frac{\sin2x}{2}\right) +c

D

−(x+sin⁡4x4+sin⁡2x2)+c-\left(x + \frac{\sin4x}{4} + \frac{\sin2x}{2}\right) +c

Answer

−(x−sin⁡4x4−sin⁡2x2)+c-\left(x - \frac{\sin4x}{4} - \frac{\sin2x}{2}\right) +c

Explanation

Solution

∫4cos⁡(x+π6)cos⁡2x⋅cos⁡(5π6+π)dx\int 4 \cos \left(x+\frac{\pi}{6}\right) \cos 2 x \cdot \cos \left(\frac{5 \pi}{6}+\pi\right) d x
=2∫(cos⁡(2x+π)cos⁡2π3)cos⁡ 2x dx=2 \int\left(\cos (2 x+\pi) \cos \frac{2 \pi}{3}\right) \cos\, 2 x\, d x
=2∫(−cos⁡2x−12)cos⁡ 2x dx=2 \int\left(-\cos 2 x-\frac{1}{2}\right) \cos\, 2 x\, d x
=∫(−2cos⁡22x−cos⁡2x)dx=\int\left(-2 \cos ^{2} 2 x-\cos 2 x\right) d x
=−∫(1+cos⁡4x+cos⁡2x)dx=-\int(1+\cos 4 x+\cos 2 x) d x
=−x−sin⁡4xx−sin⁡2x2+c=-x-\frac{\sin 4 x}{x}-\frac{\sin 2 x}{2}+c