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∫sin^2xCos^2xDx

∫sin^2xcos^2xdx=1/4∫(sin2x)^2dx=1/8∫(1-cos4x)dx=1/8(x-1/4sin4x)+C=x/8-(sin4x)/32+C满意请采纳,谢谢~

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

答案在图片上,满意请点采纳,谢谢。 愿您学业进步☆⌒_⌒☆

答: 由cos2x=1-2(sinx)^2得:(sinx)^2=1/2-cos2x/2 ∫(sinx)^2dx =∫ 1/2-cos2x/2 dx =x/2-sin2x/4 + C

∫cos2xdx = ∫(1/2)cos2xd(2x )=1/2 ∫cos2xd2x =1/2(sin 2x + C) ∫cos^2xdx=∫(cos4x +1) /2 dx =∫(cos4x +1) /8 d(4x) =1/8 ∫(cos4x +1) d(4x) =1/8 [∫cos4x d(4x)+∫1d(4x)] =1/8 [(sin 4x +C1)+ 4x+C2] =1/8(sin 4x+ 4x+C)

∫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

题干不清,无法作答。

∫cos^2xdx =∫(1+cos2x)dx/2 =∫(1+cos2x)d2x/4 =(1/4)∫[d2x+cos2xd2x] =(1/4){2x+sin2x+C1} =x/2+(sin2x)/4+C

∫cos²x dx =∫(1+cos2x)/2dx (倍半角公式) =∫(1/2)dx+(cos2x)/2dx =x/2+(sin2x)/4+c ∫cosx²dx =∫(1+cos2x)/2 dx =1/4∫(1+cos2x)d(2x) =(2x+sin2x)/4+C

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