The integral ∫ 1 - 1 3 cos x - sin x 1 + 2 3 sin 2 x d x is equal to

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

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

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

$=\int \frac{\left(\frac{\sqrt{3}}{2}-\frac{1}{2}\right)\left(\cos x-\sin x\right)}{\sin {\displaystyle \frac{\mathrm{\pi }}{3}}+\sin 2x}dx$

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

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

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

$=\frac{1}{2}\left(\int \mathrm{cosec}\left(x+\frac{\mathrm{\pi }}{6}\right)dx-\int \sec \left(x-\frac{\pi }{6}\right)dx\right)$

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

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