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sin2xDx=D()

.∫sin2xdx=1/2∫sin2xd(2x)=-1/2cos2x+C

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

d(sin2x)=2cos(2x)dx

∫(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...

y=2x dy=d(2x)=2dx 复合函数求导法则

解:原是=1/2积分sin(2x-1)d(2x-1) =-1/2cos(2x-1)+C 方法二: sin(2x-1) =sin2xcos1-cos2xsin1 原是=积分(sin2xcos1-cos2xsin1)dx =积分sin2xcos1dx-积分cos2xsin1dx =cos1x1/2积分sin2xd2x-sin1x1/2积分cos2xd2x =1/2cos1(-cos2x)-1/2sin1sin...

∵dsin2x=cos2xd2x =2cos2xdx

∫(lntanx/sin2x)dx =∫(lntanx)/2sinxcosx)dx =½∫(lntanx)cosx/(sinxcos²x)dx =½∫(lntanx)cosx/(sinx)dtanx =½∫(lntanx)/tanx)dtanx =½∫(lntanx)d(lntanx) =¼ [ln(tanx)]² + C

解: ∫(x²-1)sin2xdx =∫x²sin2xdx-∫sin2xdx =-x²(cos2x)/2 +∫xcos2x dx+∫sin2xdx =-x²(cos2x)/2 +x(sin2x)/2-1/2 ∫sin2xdx+∫sin2xdx =-x²(cos2x)/2 +x(sin2x)/2+1/2 ∫sin2xdx =-x²(cos2x)/2 +x(sin2x)/2-1/4 cos2x+C

∫[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

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