对xcos^2xdx求定积分,上限2派=360'下限0 如题

Answer 1

∫(0到2π) x*cos??x dx=∫(0到2π) x*1/2*(1+cos2x) dx=1/2*∫(0到2π) x dx+1/2*∫(0到2π) x*cos2x dx=1/2*1/2*x??(0到2π)+1/2*∫(0到2π) x*cos2x dx 对于∫(0到2π) x*cos2x dx=∫(0到2π) x*d(1/2*sin2x)=1/2*x*sin2x(0到2π)-1/2*∫(0到2π) sin2x dx=0+1/4*cos2x(0到2π)=0∴原式=1/2*1/2*x??(0到2π)+0=π??