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积分 sin^4xCos^2x Dx

方法1: 原式=∫sin⁴x cos²x =∫sin⁴x (1 - sin²x) dx =∫(sin⁴x - sin^6x) dx = ∫sin⁴x dx - ∫sin^6x dx 后面的看附图,自己整理吧 方法2: 原式=∫sin⁴x cos²x dx =∫sin²x (sinxcosx)² dx...

∵cos^4xsin^2x =cos^4x(1-cos²x) =cos^4x-cos^6x =[1+cos(2x)]²/4-[1+cos(2x)]³/8 =[1+2cos(2x)+cos²(2x)]/4-[1+3cos(2x)+3cos²(2x)+cos³(2x)]/8 =1/8+1/8cos(2x)-1/8cos²(2x)-1/8cos³(2x) =1/8+1/8cos...

(sin∧4xcos∧2x)的原函数是 ∫(sin∧4xcos∧2x)dx =∫[sin^2x(sinxcosx)^2dx =1/8∫(1-cos2x)(sin2x)^2dx =1/8∫(sin2x)^2dx-1/16∫(sin2x)^2d(sin2x) =1/16∫[1-cos4x]dx-1/48(sin2x)^3 =x/16-1/64*sin4x-1/48*(sin2x)^3+C 原函数的定义 primitive f...

=∫sin^4x(1-sin^2x)dsinx上面就是将sinx作为自变量,你可设sinx=u则:=fu^4(1-u^2)du=f[u^4-u^6]du 公式:(u^n)'=(n-1)^(n-1) ; fu^ndu=1/(n+1) *u^(n+1)+c=fu^4du-fu^6du=1/5u^5-1/7u^7+c再将u=sinx代入=1/5sin^5x-1/7sin^7x+c

原式=∫tan^4xdx =∫tan^2x*(sec^2x-1)dx =∫tan^2x*sec^2xdx-∫tan^2xdx =∫tan^2xd(tanx)-∫(sec^2x-1)dx =(1/3)*tan^3x-tanx+x+C,其中C是任意常数

得无穷。楼上那个换元换错了

∫(sinx)^2*(cosx)^4dx =(1/4)∫(sin2x)^2(1-(sinx)^2)dx =(1/4)∫(sin2x)^2(1/2+cos2x/2)dx =(1/16)∫(1-cos4x)dx+(1/16)∫(sin2x)^2dsin2x =(1/16)x-(1/64)sin4x+(1/48)(sin2x)^3+C

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