Question

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

Answer 1

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

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

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