Найти первообразную а)f(x)= sin2x cos8x б)f(x)= 2sin(в квадрате) х

[latex]\int {sin2x*cos8x} \, dx = \int {(\frac{1}{2}(sin(2x-8x)+\frac{1}{2}sin(2x+8x))} \, dx= \\ \\ = \int {(\frac{1}{2}(sin(-6x) \, dx + \int {(\frac{1}{2}(sin(10x) \, dx = \frac{1}{12}cos6x- \frac{1}{20}sin10x+C[/latex]   [latex]\int 2sin^2x \, dx=\int({1-cos2x})\, dx=x-\frac{1}{2}sin2x+C[/latex]