cos函数的四次方的定积分怎么求
三角函数的4次方积分公式？
三角函数的4次方积分公式？
(sinx)^4dx
∫[(1/2)(1-cos2x]^2dx
(1/4)∫[1-2cos2x (cos2x)^2]dx
(1/4)∫[1-2cos2x (1/2)(1 cos4x)]dx
(3/8)∫dx-(1/2)∫cos2xdx (1/8)∫cos4xdx
(3/8)∫dx-(1/4)∫cos2xd2x (1/32)∫cos4xd4x
(3/8)x-(1/4)sin2x (1/32)sin4x C
cosx的四次方的定积分怎么算？
(cosx)^4周期为π的偶函数∫(-π/2到π/2)(cosx)^4dx2∫(0到π/2)(cosx)^4dx1/2∫(cos2x 1)2dx1/2∫(cos4x 1)/2 2cos2x 1dxsin4x/16 sin2x/2 3x/43π/8或直接牛-莱公式2*3/4*1/2*π/23π/8
微积分问题，sinx的4次方的积分如何求？
(1)先求S(sin2x)^2dxx/2-(1/4)sina2xcos2xS(sin2x)^2dx-(1/2)Ssin2xdcos2x-(1/2){sin2xcos2x-2Scos2xcos2xdx}-(1/2){sin2xcos2x-2S(1-(sinx)^2)dx}-(1/2)sin2xcos2xS(1-(sinx)^2)dx-(1/2)sin2xcos2xx-S((sinx)^2)dxS(sin2x)^2dx-(1/4)sin2xcos2x(1/2)x(2)S(sinx)^4dx-S(sinx)^3dcosx-(sinx)^3cosx3Scos^2xsin^2xdx-(sinx)^3cosx(3/4)S4cos^2xsin^2xdx-(sinx)^3cosx(3/4)Ssin^22xdx-(sinx)^3cosx(3/4)(x/2-(1/4)sina2xcos2x)下面自己整理吧。
cosx四次方的定积分公式？
(cosx)^4周期为π的偶函数∫(-π/2到π/2)(cosx)^4dx2∫(0到π/2)(cosx)^4dx1/
2∫(cos2x 1)2dx1/
2∫(cos4x 1)/2 2cos2x 1dxsin4x/16 sin2x/2 3x/43π/8或直接牛-莱公式2*3/4*1/2*π/23π/8