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求X(sinx)的平方 dx的不定积分
人气:170 ℃ 时间:2019-10-08 16:53:01
解答
x(sinx)^2=x*(1-cos2x)/2=1/2*x-1/2xcos2x
∫x(sinx)^2dx=1/2∫xdx- 1/2∫xcos2xdx
=1/4x^2 -1/4∫xdsin2x
=1/4x^2-1/4(xsin2x-∫sin2xdx)
=1/4x^2-1/4xsin2x+1/4∫sin2xdx
=1/4x^2-1/4xsin2x-1/8cos2x+c
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