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∫sin4xCos2xDx 积分

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∫(sinx)^2*(cosx)^4dx =(1/4)∫(sin2x)^2(1-(sinx)^2)dx =(1/4)∫(sin2x)^2(1/2+cos2x/2)dx =(1/16)∫(1-cos4x)dx+(1/16)∫(sin2x)^2dsin2x =(1/16)x-(1/64)sin4x+(1/48)(sin2x)^3+C

∵cos^4xsin^2x =cos^4x(1-cos²x) =cos^4x-cos^6x =[1+cos(2x)]²/4-[1+cos(2x)]³/8 =[1+2cos(2x)+cos²(2x)]/4-[1+3cos(2x)+3cos²(2x)+cos³(2x)]/8 =1/8+1/8cos(2x)-1/8cos²(2x)-1/8cos³(2x) =1/8+1/8cos...

∫sin2xdx =(1/2)∫sin2xd(2x) =-(1/2)cos2x + C

∫(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

-((3 x)/2) + 1/4 Sin[2 x] + Tan[x]

应该是∫(sinx)^2cos2xdx,用降幂公式把原式打开即可,解法如下:

这个有公式的

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