✔ 最佳答案
w=∫(sinx*cosx)^2*cos^2x*dx=∫[sin(2x)]^2/4*[cos(2x)+1]/2*dx=1/8*∫[sin(2x)^2*cos(2x)+sin(2x)^2]dx=1/16*∫sin(2x)^2*d[sin(2x)]+1/8*∫[1-cos(4x)]dx/2=sin(2x)^3/48+1/16*∫dx-1/16*∫cos(4x)dx=[sin(2π)^3-sin(0)^3]/48+x/16-sin(4x)/64=0+(π-0)/16-[sin(4π)-sin(0)]/64=π/16+0=π/16..........ans