Найдите первообразную для функции: ответ должен быть 1/2 tg(2x+ pi/3)+ c

Решите задачу:

$f(x)=\frac{sin(\frac{\pi}{6}-2x)}{cos^3(\frac{\pi}{3}+2x)}=\frac{sin(\frac{\pi}{2}-\frac{\pi}{3}-2x)}{cos^3(\frac{\pi}{3}+2x)}=\Big [\; sin(\frac{\pi}{2}-a)=cosa\, \Big ]=\frac{sin(\frac{\pi}{2}-(\frac{\pi}{3}+2x))}{cos^3(\frac{\pi}{3}+2x)}=\\\\=\frac{cos(\frac{\pi}{3}+2x)}{cos^3(\frac{\pi}{3}+2x)}=\frac{1}{cos^2(\frac{\pi}{3}+2x)}}\; ;\\\\\\F(x)=\int f(x)\, dx=\int \frac{sin(\frac{\pi}{6}-2x)}{cos^3(\frac{\pi}{3}+2x)}\, dx=\int \frac{dx}{cos^2(\frac{\pi}{3}+2x)}=\frac{1}{2}\cdot tg(\frac{\pi}{3}+2x)+C\; .$

$\star \; \; \int \frac{dx}{cos^2(kx+b)}=\frac{1}{k}\cdot tg(kx+b)+C\; \; \star$

$\star \; \; \frac{\pi}{6}=\frac{\pi}{2}-\frac{\pi}{3}\; \; \star$