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∫sinxsin2xsin3xDx

这题利用公式求

积化和差∫sinxsin2xsin3xdx=1/2∫(cosx-cos3x)sin3xdx=1/2∫cosxsin3xdx-1/2∫cos3xsin3xdx=1/4∫(sin2x+sin4x)dx-1/4∫sin6xdx=-1/8cos2x-1/16cos4x+1/24cos6x+C数学软件验算:

sinx sin2x sin3x =0.5(cos3x - cosx)sin3x =0.5sin3x cos3x -0.5sin3xcosx =0.25 sin6x -0.25(sin4x+sin2x) 看起来你符号正好反了

被积函数是奇函数,而积分区间关于原点对称,根据”奇函数在对称区间的积分为0“可知,原式=0.

证明:由题意,P=2xcosy-y2sinx,Q=2ycosx-x2siny,在整个平面上具有一阶连续偏导数,且?P?y=?2xsiny?2ysinx=?Q?x∴曲线积分I与积分路径无关.取路径从(0,0)到(2,0)再到(0,3),则I=∫202xdx+∫30(2ycos2?4siny)dy=4+9cos2+4cos3-4=9cos...

用分步积分 ∫(xcosx)/(sinx)^3 dx =∫(x)/(sinx)^3 dsinx =-1/2∫(x) d(1/sin^2x) =-1/2s/sin^2x+1/2∫1/sin^2xdx =-1/2s/sin^2x+1/2∫csc^2xdx =-1/2s/sin^2x-1/2cotx+C

∫[0,π]x^2(sinx^2)dx =∫[0,π]x^2(1/2)(1-cos2x)dx =∫[0,π](1/2)x^2dx-∫[0,π](1/2)x^2cos2xdx =(1/6)π^3 -(1/4)∫[0,π]x^2dsin2x =(1/6)π^3+(1/2)∫[0,π]xsin2xdx =(1/6)π^3+(-1/4)∫[0,π]xdcos2x =(1/6)π^3+(-1/4)π+(1/4)∫[0,π]cos2xdx =(1/6)π^3+(...

∫ cos²x dx = ∫ (1 + cos2x)/2 dx = x/2 + (1/4)sin4x + C ∫ cos³x dx = ∫ (1 - sin²x) dsinx = sinx - (1/3)sin³x + C ∫ cos⁴x dx = ∫ (cos²x)² dx = ∫ [(1 + cos2x)/2]² dx = (1/4)∫ (1 + 2cos2x + ...

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