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sin2xDx=D()

公式:df(x) = f'(x)dx (sin2x)' = (2x)' * cos2x = 2cos2x dsin2x = 2cos2xdx 你大概有个笔误吧。

.∫sin2xdx=1/2∫sin2xd(2x)=-1/2cos2x+C

答: 原积分 =-∫1/2×sin2xd(2x) =-1/2×(-cos2x) + C =cos2x/2 + C

∫(x²-1)sin2xdx =∫x²sin2xdx-∫sin2xdx =-1/2∫x²dcos2x+1/2cos2x =-1/2x²cos2x+1/2∫cos2xdx²+1/2cos2x =-1/2x²cos2x+∫xcos2xdx-1/2cos2x =-1/2x²cos2x+1/2∫xdsin2x-1/2cos2x =-1/2x²cos2x+1/2xsin2x-1/2...

∫sin2xdx =(1/2)∫sin2xd(2x) =-(1/2)cos2x + C

∫ cos2x dx = (1/2)∫ cos2x d(2x) = (1/2)sin2x + C (1/2)∫ sin2x dx = (1/4)∫ sin2x d(2x) = (1/4)(-cos2x) + C = -(1/4)cos2x + C ∴∫ cos2x dx ≠ (1/2)∫ sin2x dx

∫[0,π/2]sin2x dx =(1/2)∫[0,π/2]sin2x d(2x) =-(cos2x)/2 |[0,π/2] =-(cosπ)/2+(cos0)/2 =1/2+1/2 =1

y=2x dy=d(2x)=2dx 复合函数求导法则

注意 (cos²x)'= 2cosx *(cosx)'= -2sinx *cosx= -sin2x 所以就得到 ∫sin2xf'(cos²x)dx = - ∫f '(cos²x)d(cos²x) = -f(cos²x) +C,C为常数

=1/2∫xsin2xd2x =-1/2∫xdcos2x =-1/2*xcos2x+1/2∫cos2xdx =-1/2*xcos2x+1/4∫dsin2x =-1/2*xcos2x+1/4*sin2x+C

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