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∫[sin^3(x)]CosxDx等于多少?求过程

∫[sin^3(x)]cosxdx=∫[sin^3(x)]d(sinx)=sin^4(x)/4+C

变换加换元如下

1:∫sin³xcosxdx=∫sin³xdsinx=1/4(sin^4x)+C2:∫(x²+e²)dx=1/3(x³)+e²x+C

你好!如下图先用积化和差,再凑微分,第二图是积化和差公式。经济数学团队帮你解答,请及时采纳。谢谢!

(x+sin(x))/(4cos(x)-4) 首先改写降次: xcos^4(x/2)/sin^3(x) = 1/8 xcot(x/2)csc^2(x/2) ... 注:csc(x)=1/sin(x) 换元积分:令u=x/2: dx=2du, ∫ 1/8 xcot(u)csc^2(u) 2du =1/8 * 2 * 2 ∫ ucot(u)csc^2(u) du 然后分部积分, =1/2 u(-1/2cot^2...

∫[0,π/2] sin^3 xcosxdx =∫[0,π/2] sin^3 xdsinx =sin^4x/4[0,π/2] =1/4

此题用罗必塔法则比较简单,lim(x--0) f(x)/g(x)=lim(x--0)f'(x)/g'(x) =lim(x--0)[sin(sinx)^2]cosx/(3x^2+4x^3) =(等价无穷小代换) lim(x--0)x^2/3x^2=1/3,所以选B.

原式=∫sin²x(sinxdx) =∫(1-cos²x)(-dcosx) =∫(cos²x-1)dcosx =cos³x/3-cosx+C

∫(cosx)^5dx =∫(cosx)^4 cosdx = ∫ (1-sin^2 x)^2 d(sinx) =∫[1-2(sinx)^2+(sinx)^4]d(sinx) =sinx-2/3*(sinx)^3+1/5*(sinx)^5+C ∫sin3xsin5xdx =1/2*∫[ cos2x-cos8x]dx =1/2*∫ cos2xdx-1/2*∫cos8xdx =1/4∫ cos2xd(2x)-1/16∫ cos8xd(8x) =1/4*si...

∫ cos²x/sin³x dx = ∫ cot²x * cscx dx = ∫ (csc²x - 1) * cscx dx = ∫ csc³x dx - ∫ cscx dx = ∫ csc³x dx - ln|cscx - cotx| 记A = ∫ csc³x dx = ∫ cscx * csc²x dx = ∫ cscx d(- cotx) = - cscxcotx...

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