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∫x/(1+Cos²x)Dx

cos2x=2cos^2x-1 所以 1+cos2x=2cos²x

=∫xdsin(x+1) =xsin(x+1)-∫sin(x+1)dx =xsin(x+1)+cos(x+1)+C

∫xcos(1/x)dx=½ ∫2xcos(1/x)dx=½ x²cos(1/x)-½ ∫sin(1/x)dx 令u=1/x,则du=-xˉ²dx=-1/x²dx ,则dx=-xˉ²du=-1/u²du ∫sin1/xdx=∫sinu(-1/u²)du=∫sinud(1/u) 用分部积分法: ∫sin1/xdx=∫sinu(-1/u²...

(tanx)'=1/cos²x ∴∫(1/cos²x)dx=tanx+C

∫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

显然d(1/x)= -1/x² dx 所以得到 原积分 =∫ (1/x²) *cos(1/x) dx =∫ -cos(1/x) d(1/x) = -sin(1/x) +C,C为常数

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 ...

把log改成ln就可以了

∫ x/(1+cosx) dx =∫ x/[2cos²(x/2)] dx =∫ xsec²(x/2) d(x/2) =∫ x dtan(x/2) 分部积分 =xtan(x/2) - ∫ tan(x/2) dx =xtan(x/2) - 2∫ sin(x/2)/cos(x/2) d(x/2) =xtan(x/2) + 2∫ 1/cos(x/2) dcos(x/2) =xtan(x/2) + 2ln|cos(x/2)| + C...

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