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∫从Cosx/sin^3(x)Dx的不定积分

变换加换元如下

怎么问到这里来的? 先用积化和差公式 cosxsin3x=(1/2)[sin4x-sin(-2x)] =(1/2)(sin4x+sin2x) =(sin4x/2)+(sin2x)/2 ∴∫cosxsin3xdx =∫[(sin4x/2)+(sin2x)/2]dx =(1/2)∫sin4xdx+(1/2)∫sin2xdx =(1/2)(1/4)∫sin4xd(4x)+(1/2)(1/2)∫sin2xd(2x) =(1/8...

∫(1-sinx^2)d(sinx)=sinx-1/3sinx^3

(cos(x))^3*dx=(cos(x))^2*cosxdx=[1-(sin(x))^2]d(sinx(x))==> inf[(cos(x))^3,x]=sin(x)-(sin(x))^3/3+C

令 x/2=t x=2t dx=2dt ∫cosxsin(x/2)dx=∫2cos(2t)sintdt=2∫-cos(2t)dcost=2∫-(2cos^2t-1)dcost=2∫-2cos^2t+1dcost=2(-2cos^3t/3+cost)+C =-4cos^2t/3+2cost+C=-4cos^3(x/2)/3+2cos(x/2)+C

是cosx^4,还是(cosx)^4?

主要是在化简指数问题上,尽量把次方形式变为复角形式,例如cos(nx)和sin(nx)等,比较好积 cos^6x = (cos²)³ = [(1+cos2x)/2]³ = (1/8)(1+cos2x)³ = (1/8)(1+3cos2x+3cos²2x+cos³2x) = (1/8)+(3/8)cos2x+(3/8)(1/...

∫dx/(3+sin^2x)=∫dx/(4-cos^2x) ∫dx/[2+cosx)][2-cosx]= ∫(1/4)/[2-cosx]+(1/4)/[2+cosx]dx= (1/4)∫1/[2-cosx]+1/[2+cosx]dx= (1/4){{2/3^(1/2)arctan{[(2+1)/(2-1)]^(1/2)tan(x/2)}}+{2/3^(1/2)arctan{[(2-1)/(2+1)]^(1/2)tan(x/2)}}}

∫sin⁴*cos³x dx = ∫sin⁴*cos²x dsinx = ∫sin⁴*(1-sin²x) dsinx = ∫(sin⁴x - sin^6x) dsinx = (1/5)[sinx]*5 - (1/7)[sinx]^7 + C 希望对你有帮助

因为sin2x = 2sinxcosx; ∫sin2xcosxdx = ∫2sinxcosxcosxdx = -2∫cosx^2dcosx = -2/3∫cosx^3

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