The integral ∫ (1 - 1 3 )( x - x) (1 + 2 3 2x) dx is equal to

Let $I = ∫ \dfrac{\left(1 - \dfrac{1}{\sqrt{3}}\right)\left(\cos x - \sin x\right)}{\left(1 + \dfrac{2}{\sqrt{3}}\sin 2x\right)}dx$

Multiplying by $\dfrac{\sqrt{3}}{2}$ in numerator and denominator, we get

$I = ∫ \dfrac{\left(\dfrac{\sqrt{3}}{2} - \dfrac{1}{2}\right)\left(\cos x - \sin x\right)}{\left(\dfrac{\sqrt{3}}{2} + \sin 2x\right)}dx$

$= ∫ \dfrac{\left(\dfrac{\sqrt{3}}{2} - \dfrac{1}{2}\right)\left(\cos x - \sin x\right)}{sin\dfrac{\pi}{3} + \sin 2x}dx$

$= ∫ \dfrac{\left(\dfrac{\sqrt{3}}{2}\cos x - \dfrac{1}{2}\cos x - \dfrac{\sqrt{3}}{2}\sin x + \dfrac{1}{2}\sin x\right)}{2\sin \left(x + \dfrac{\pi}{6}\right)\cos \left(x - \dfrac{\pi}{6}\right)}dx$

$= ∫ \dfrac{\left(\cos \left(x - \dfrac{\pi}{6}\right) - \sin \left(x + \dfrac{\pi}{6}\right)\right)}{2\sin \left(x + \dfrac{\pi}{6}\right)\cos \left(x - \dfrac{\pi}{6}\right)}dx$

$= \dfrac{1}{2}\left(∫ \dfrac{dx}{\sin \left(x + \dfrac{\pi}{6}\right)} - ∫ \dfrac{dx}{\cos \left(x - \dfrac{\pi}{6}\right)}\right)$

$= \dfrac{1}{2}\left(∫ cosec\left(x + \dfrac{\pi}{6}\right)dx - ∫ \sec \left(x - \dfrac{\pi}{6}\right)dx\right)$

$= \dfrac{1}{2}\left(\ln \left|\tan\dfrac{x + \dfrac{\pi}{6}}{2}\right|\right) - \dfrac{1}{2}\left(\ln \left|\tan\dfrac{x - \dfrac{\pi}{6}}{2} + \dfrac{\pi}{4}\right|\right) + C$

$= \dfrac{1}{2}\ln \left|\dfrac{\tan \left(\dfrac{x}{2} + \dfrac{\pi}{12}\right)}{\tan \left(\dfrac{x}{2} + \dfrac{\pi}{6}\right)}\right| + C$