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∫Cos2x/Cosx^2sinx^2Dx不定积分多少

∫(cos2x/[(cosx)^2.(sinx)^2] )dx =4∫[cos2x/(sin2x)^2]dx =2∫d(sin2x)/(sin2x)^2 =-2/(sin2x) + C

∫cos2x/cosx^2sinx^2dx =2∫cos2x/(sin2x)^2dx =∫1/(sin2x)^2dsin2x =-1/sin2x+C

I = ∫e^x(sinx)^2dx = (1/2)∫e^x(1-cos2x)dx = (1/2)e^x - (1/2)∫e^xcos2xdx 其中 J = ∫e^xcos2xdx = ∫cos2xde^x = e^xcos2x + 2∫sin2xe^xdx = e^xcos2x + 2e^xsin2x - 2∫cos2xe^xdx = e^x(cos2x + 2sin2x) - 2J, 则 J = (1/3)e^x(cos2x + 2sin2...

∫cos²xdx=∫[(1+cos2x)/2]dx=(1/2)[∫dx+∫cos2xdx] =(1/2)[x+(1/2)∫cos2xd(2x)]=(1/2)[x+(1/2)sin2x]+c

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

∫(1+sinx)cos²xdx =∫cos²xdx+∫sinxcos²xdx =∫(1+cos2x)/2dx-∫cos²xdcosx =1/2x+1/4sin2x-1/3cos³xdx+C

由题意可知:∫exsinx2dx=12∫ex(1?cos2x)dx=12ex?12∫excos2xdx ∫excos2xdx=excos2x+2∫exsin2xdx =excos2x+2exsin2x-4∫excos2xdx∴∫excos2xdx=15ex(cos2x+2sin2x)+C 代入得:原式=12ex?110ex(cos2x+2sin2x)+C

参考答案:

∫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∫(cosx)^2 dx =∫ 1/2 *[2(cosx)^2 -1] +1/2 dx =∫ 1/2 *cos2x +1/2 dx =1/4 *cos4x +x/2 +C 而∫ (cosx)^3 dx =∫ (cosx)^2 d(sinx) =∫1-(sinx)^2 d(sinx) =sinx -1/3 *(sinx)^3 +C

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