Question

Найдите первообразную для функции:$f(x)=\frac{sin(\frac{\pi }{6}-2x )}{cos^{3} (\frac{\pi }{3}+2x) }$

Answer 1

$\frac{\pi}{3} = \frac{\pi}{2}-\frac{\pi}{6}\\\\\int\limits \frac{sin(\frac{\pi}{6}-2x)}{cos^3(\frac{\pi}{3}+2x)} \, dx = \int\limits \frac{sin(\frac{\pi}{6}-2x)}{cos^3(\frac{\pi}{2}-\frac{\pi}{6}+2x)} \, dx = \int\limits\frac{sin(\frac{\pi}{6}-2x)}{cos^3(\frac{\pi}{2}-(\frac{\pi}{6}-2x))} \, dx =\\\\= \int\limits\frac{sin(\frac{\pi}{6}-2x)}{sin^3(\frac{\pi}{6}-2x)} \, dx = \int\limits\frac{dx}{sin^2(\frac{\pi}{6}-2x)} = -\frac{1}{2}\int\limits\frac{d(\frac{\pi}{6}-2x)}{sin^2(\frac{\pi}{6}-2x)} =$

$= \frac{1}{2}ctg(\frac{\pi}{6}-2x) + C$

Ответ: $\frac{1}{2}ctg(\frac{\pi}{6}-2x) + C$