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sin5xCos2x的不定积分

sin(7x)=sin5xcos2x+cos5xsin22∫sin7xdx=∫sin5xcos2xdx+∫cos5xsin2xdx因为∫cos5xsin2xdx=1/5*[sin5xsin2x-2∫sin5xcos2xdx]=1/5*sin5xsin2x-2/5*∫sin5xcos2xdx所以3/5*∫sin5xcos2xdx+1/5*sin5xsin2x=∫sin7xdx∫sin7xdx=-1/7*cos7x所以∫sin5xcos2xdx=[-1/7*cos7x-1/5*sin5xsin2x]*5/3

应该是∫(sinx)^2cos2xdx,用降幂公式把原式打开即可,解法如下:

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

积化和差原式=∫1/2[sin(5x+2x)+sin(5x-2x)]dx=1/2∫sin7xdx+1/2∫sin3xdx=1/14∫sin7xd7x+1/6∫sin3xd3x=-1/14*cos7x-1/6*cos3x+C

∫sin5xcos2xdx = 1/2*∫sin5xcos2xd(2x)=-1/2*∫sin5xd(sin2x)

利用三角函数和差化积公式.∫sin3xcos2xd(x)=1/2∫(sin5x+sinx)dx=1/2(∫sin5xdx+∫sinxdx)=-1/10cos5x-cosx+C

∫sin3xcos2xdx=∫[(sin5x+sinx)/2]dx==0.5∫sin5xdx+0.5∫sinxdx==0.5/5(-con5x)+0.5(-conx)+C==-0.1con5x-0.5conx+C.

sin5x=Sin3xCos2x+Sin2xCos3x sinx=Sin3xCos2x-Sin2xCos3x ∫Sin3xCos2xdx=0.5 ∫Sin5x+Sinxdx=-0.1cos5x-0.5cosx

∫ e^xsinx dx=(1/2)∫ e^x(1-cos2x) dx=(1/2)e^x - (1/2)∫ e^xcos2x dx (1) 下面计算:∫ e^xcos2x dx=∫ cos2x d(e^x) 分部积分=e^xcos2x + 2∫ e^xsin2x dx=e^xcos2x + 2∫ sin2x d(e^x) 再分部积分=e^xcos2x + 2e^xsin2x - 4∫ e^xcos2x dx 将 -4∫ e^xcos2x

先做积化和差:2sin3x*cos2x=sin5x+sinx下面就简单了,直接化为正弦函数的积分

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