Question

cos2x^(1/3)*sin2x dx найти интеграл

Answer 1

$\int (cos2x)^{1/3}sin2x\, dx=[\, t=cos2x\; ,\; dt=-2sin2x\, dx\, ]=\\\\=-\frac{1}{2} \int t^{1/3}\cdot dt=- \frac{1}{2}\cdot \frac{t^{4/3}}{4/3}+C=- \frac{3}{8}\cdot \sqrt[3]{cos^42x}+C$