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∫1/(sin^2xCos^2x)

(sinx*cosx)^2=0.25*sin(2x)^2 积分=-2/sin(2*x)*cos(2*x)+C

答案在图片上,满意请点采纳,谢谢。 愿您学业进步☆⌒_⌒☆

利用半角公式如图降次计算。经济数学团队帮你解答,请及时采纳。谢谢!

sin²xcos²x =(1/4)(2sinxcosx)² =sin²(2x)/4 =[1-cos(2x)]/8 =1/8 -cos(2x)/8

=∫(cos^2x-sin^2x)/(cos^2xsin^2x)=∫(1/sin^2x - 1/cos^2x)=∫csc^2x-∫sec^2x=cotx-tanx

∫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

利用两角和的余弦公式cos2X = cos(X+X) = cosXcosX- sinXsinX = cos^2 X- sin^2 X = 2cos^2 X- 1=1 - 2sin^2 X

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

∫(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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