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求 ∫(x^+2)sinxDx 的不定积分 尽可能的详细点 谢谢

∫(x^2)sinxdx = - ∫x^2dcosx = -x^2cosx + ∫2xdcosx= -x^2cosx + 2∫xdsinx = -x^2cosx + 2xsinx - ∫sinxdx= -x^2cosx + 2xsinx + cosx + C

∫xsinxdx=-∫xdcosx=-xcosx+∫cosxdx=-xcosx+2∫xcosxdx=-xcosx+2∫xdsinx=-xcosx+2xsinx-2∫sinxdx=-xcosx+2xsinx+2cosx+c

∫x^2.sin2xdx=(-1/2)∫x^2dcos2x=(-1/2)x^2.cos2x +∫x.cos2xdx=(-1/2)x^2.cos2x +(1/2)∫x.dsin2x=(-1/2)x^2.cos2x +(1/2)x.sin2x -(1/2)∫sin2x dx=(-1/2)x^2.cos2x +(1/2)x.sin2x +(1/4)cos2x + C

cos2x=1-2sin(x^2)则:∫sin(x^2)dx=∫(1-cos2x)/2dx =∫1/2dx-∫1/2cos2xdx =1/2x-1/4∫cos2xd2x =1/2x-1/4sin2x

∫sin(x/2)dx=1/2*∫2sin(x/2)dx=1/2∫(1-cosx))dx=1/2*x-1/2∫cosxdx=1/2*x -1/2*sinx +c

原式=∫sinx/2sinxcosx dx=∫1/2cosxdx=1/2*∫secxdx=1/2*ln|secx+tanx|+C 说明:∫secxdx=ln|secx+tanx|+C 是常见公式,要记住的.

x(sinx)^2=x/2(1-cos2x) 然后用分步积分法就可以了

∫x^2sin2xdx=-1/2∫x^2d(cos2x)=-1/2[cos2x*x^2-∫2x*cos2xdx]=-1/2[cos2x*x^2-∫xd(sin2x)]=-1/2[cos2x*x^2-(sin2x*x-∫sin2xdx)]=-1/2cos2x*x^2+1/2sin2x*x-1/2∫sin2xdx=-1/2cos2x*x^2+1/2sin2x*x+1/4cos2

解:原式= -1/2∫xdcos2x = -1/2(xcos2x - ∫xdsin2x) = -1/2(xcos2x - xsin2x + ∫sin2xdx) =[(1-2x)cos2x + 2xsin2x]/4 + C. 你的第一步就写错了,应该是:原式=1/2∫xsin2xd(2x) 你的第二步也写错了,应该是:=1/2[x(-cos2x) - ∫2x(-cos2x)dx] 所以你的结果就错了.

sinxcosx

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