∫(xsin x)²dx 不定积分怎么求
人气:260 ℃ 时间:2020-01-30 07:40:11
解答
∫(xsin x)²dx =Sx^2*(sinx)^2 dx=Sx^2*(1-cos2x)/2 dx=1/2*Sx^2dx-1/2*Sx^2 cos2x dx=1/6*x^3-1/4*Sx^2dsin2x=1/6*x^3-1/4*x^2sin2x+1/4*Ssin2xdx^2=1/6*x^3-1/4*x^2sin2x+1/2*Sxsin2xdx=1/6*x^3-1/4*x^2sin...
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