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∫(1+sin2x)^3Cos2xDx, 区间是 {0,π/4},求定积分...

原式=0.5∫x^2 (1-cos2x)dx =0.5∫x^2dx-0.5∫x^2cos2xdx =0.5x^3/3-0.5[ x^2 *0.5sin2x-∫xsin2xdx] =0.5x^3/3-0.25x^2sin2x+0.5[-0.5xcos2x+∫0.5cos2xdx] =[0.5x^3/3-0.25x^2sin2x-0.25xcos2x+0.125sin2x] =[0.5π^3/3-0.25π]-[0] =π^3/6-π/4

∫cosxcos2xdx =∫(1-2sin²x)d(sⅰnx) =sⅰnx-(2/3)sⅰn³x+C

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