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 + ...