∫(xsin x)²dx 不定积分怎么求
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∫(xsin x)²dx 不定积分怎么求
![∫(xsin x)²dx 不定积分怎么求](/uploads/image/z/15012451-19-1.jpg?t=%E2%88%AB%EF%BC%88xsin+x%EF%BC%89%26%23178%3Bdx+%E4%B8%8D%E5%AE%9A%E7%A7%AF%E5%88%86%E6%80%8E%E4%B9%88%E6%B1%82)
∫(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^2sin2x-1/4*Sxdcos2x
=1/6*x^3-1/4*x^2sin2x-1/4*xcos2x+1/4*Scos2xdx
=1/6*x^3-1/4*x^2sin2x-1/4*xcos2x+1/8*sin2x+c
=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^2sin2x-1/4*Sxdcos2x
=1/6*x^3-1/4*x^2sin2x-1/4*xcos2x+1/4*Scos2xdx
=1/6*x^3-1/4*x^2sin2x-1/4*xcos2x+1/8*sin2x+c