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∫sin2xCos3xDx

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

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

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

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

∫sin2xcos3xdx=(cosx)/2-(cos5x)/10+C.(C为积分常数)∫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扩展资料:常用积分公式:1)∫0dx=c 2)∫x^udx=(x^(u+1))/(u+1)+c3)∫1/

∫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.

原式=∫1dx-∫sinx*sinxdx=x+∫(1-cosx)dcosx=x+cosx-cosx/3+c

∫sin2xcos3xdx=0.5∫sin5x-0.5∫sinx=-0.1cos5x+0.5cosx+c利用sinA*cosB=0.5sin(A+B)+0.5sin(A-B)

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