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∫x/1%Cos2x Dx=?

cos2x=cos²x-sin²x=2cos²x-1=1-2sin²x 所以1-cos2x=1-(1-2sin²x)=2sin²x 原式=∫x/2sin²x dx =1/2*∫x/sin²xdx =1/2*∫xcsc²xdx =-1/2*∫xdcotx =-1/2*xcotx+1/2*∫cotxdx =-1/2xcotx+1/2∫cosx/sinxdx ...

∫xcos2xdx =(1/2)∫xdsin2x =(1/2)x.sin2x -(1/2)∫sin2xdx =(1/2)x.sin2x +(1/4)cos2x + C

cos2x=cos²x-sin²x=2cos²x-1=1-2sin²x 所以1-cos2x=1-(1-2sin²x)=2sin²x 原式=∫x/2sin²x dx =1/2*∫x/sin²xdx =1/2*∫xcsc²xdx =-1/2*∫xdcotx =-1/2*xcotx+1/2*∫cotxdx =-1/2xcotx+1/2∫cosx/sinxdx ...

cos2x=2cos^2x-1 所以 1+cos2x=2cos²x

∫xcos²xdx=∫x(1+cos2x)/2dx=1/2(∫xdx+∫xcos2xdx) =1/2(1/2x²+∫xcos2xdx) =1/2(1/2x²+1/2∫xdsin2x) =1/2(1/2x²+1/2(xsin2x-∫sin2xdx)) =1/2(1/2x²+1/2xsin2x+1/4cos2x)+C

这里就是用来凑微分的办法, 显然求导得到(cos2x)'= -2sin2x 所以就有 d(cos2x)= -2sin2x dx 于是就得到了 ∫ x sin2xdx= -1/2 *∫ xd(cos2x)

∫(x-1)cos2xdx =∫(xcos2x-cos2x)dx =∫xcos2xdx-∫cos2xdx =1/2∫xdsin2x-∫cos2xdx =1/2(xsin2x-∫sin2xdx)(分部积分法)-∫cos2xdx =1/2xsin2x-1/2∫sin2xdx-∫cos2xdx =1/2xsin2x-1/4∫sin2xd2x-∫cos2xdx =1/2xsin2x+1/4cos2x-1/2sin2x 你是第三步错了...

∫[0→π/2] 2xcos2x dx =∫[0→π/2] x d(sin2x) =xsin2x - ∫[0→π/2] sin2x dx =xsin2x + (1/2)cos2x |[0→π/2] =-(1/2) - (1/2) =-1 希望可以帮到你,不明白可以追问,如果解决了问题,请点下面的"选为满意回答"按钮,谢谢。

显然(cosx)^2=1/2cos2x+1/2 所以∫ (cosx)^4 dx =∫ 1/4(cos2x+1)^2 dx =∫ 1/4 *(cos2x)^2 +1/2 *cos2x +1/4 dx =∫ 1/8 *cos4x +1/2 *cos2x +3/8 dx = -1/32 *sin4x -1/4 *sin2x +3x/8 代入上下限π/2和0 得到积分值=3π/16

1+cos2x=2(cosx)^2 =∫1/2(cosx)^2 dx =0.5 *∫1/(cosx)^2 dx =0.5tanx +C,C为常数

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