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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

非初等,积不出来的

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...

∫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

∫(sinx^2)dx =∫(1-cos2x)/2dx =∫1/2dx-½∫(cos2x/2)dx+c =x/2 -∫(cos2x/4)d2x+c =x/2-sin2x/4+c

∫sin²dx=∫1-cos²dx =∫1-(1+cos2x)/2dx =∫1/2-1/2cos2xdx =(1/2)x-(1/2)∫cos2xdx =(1/2)x-(1/4)sin2x+c(c为任意常数)

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

S(sinx)^4(cosx)^2dx=1/8*S(sin2x)^2 *(1-cos2x)dx =1/8*S(sin2x)^2dx-1/8*S(sin2x)^2cos2xdx =1/16*S(1-cos4x)dx-1/16*S(sin2x)^2dsin2x =1/16*x-1/16*Scos4xdx-1/48*(sin2x)^3 =1/16*x-1/64*sin2x-1/48*(sin2x)^3+c 积分区间是0到二分之...

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

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