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∫sin∧2xCos∧3xDx

cos3x=∫sin2xcos3xdx=∫1/2(sin(2x+3x)+sin(2x-3x))dx=1/2∫sin5xdx-1/2∫sinxdx=1/10∫sin5xd5x+1/2∫dcosx=(cosx)/2-(cos5x)/10+C

解:∫sin^3xcos^2xdx =-∫sin^2xcos^2xdcosx =-∫(1-cos^2x)*cos^2xdcosx =-∫(cos^2x-cos^4x)dcosx =(1/5)*cos^5x-(1/3)*cos^3x

解: ∫sin^2xcos^3xdx =∫(sinx)^2cosx(1-(sinx)^2)dx =∫(sinx)^2cosxdx-∫(sinx)^4cosxdx =∫(sinx)^2d(sinx)-∫(sinx)^4d(sinx) =(sinx)^3/3-(sinx)^5/5+C

用积化和差先展开有:sin2xcos3x=1/2(sin5x_sinx),积分得1/2cosx_1/10cos5x注:_为减号

∫sin^2x cos^2x dx =∫1/4*(sin2x)^2 dx =∫1/4*(1-cos4x)/2 dx =1/8*∫(1-cos4x)dx =1/8*(x-1/4*sin4x+c) =x/8-sin4x/32+c

你好!=sin^3xcos^2x希望对你有所帮助,望采纳.

先利用积化和差公式:sin(A)cos(B) = (1/2)[sin(A+B) + sin(A-B)]∴sin(2x)cos(3x) = (1/2)[sin(2x+3x) + sin(2x-3x)]= (1/2)[sin5x + sin(-x)]= (1/2)(sin5x - sinx)∴∫ sin(2x)cos(3x) dx= (1/2)∫ sin(5x) dx - (1/2)∫ sinx dx= (1/2)(1/5)∫ sin(5x) d(5x) - (1/2)∫ sinx dx= (1/10)(-cos(5x)] + (1/2)cosx + C= (1/10)[5cosx - cos(5x)] + C

∫sin^2xcos^3xdx =∫(sinx)^2cosx(1-(sinx)^2)dx =∫(sinx)^2cosxdx-∫(sinx)^4cosxdx =∫(sinx)^2d(sinx)-∫(sinx)^4d(sinx) =(sinx)^3/3-(sinx)^5/5+C

∫sin^2xcos^4xdx=∫(1-cos^2x)cos^4xdx=∫cos^4xdx-∫cos^6xdx 利用积分表得到.

∫sin2xcos3xdx=∫1/2[sin(2x+3x)+sin(2x-3x)]dx=1/2∫sin5xdx-1/2∫sinxdx=1/10∫sin5xd5x+1/2∫dcosx=(cosx)/2-(cos5x)/10+C你好,很高兴为你解答,希望对你有所帮助,若满意请及时采纳.

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