7、∫0到π/2 cos^4xdx=?

1个回答

原式=0.5∫(1+cos2x)^2 dx=0.5∫[1+2cos2x+(cos2x)^2]dx
=0.5∫dx+0.5∫cos2xd(2x)+0.25∫[1+cos4x]dx
=[0.5x+0.5sin2x+0.25x+0.25/4*sin4x]0到π/2
=0.75*π/2
=3π/8