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∫xsin^2xdx=?
人气:344 ℃ 时间:2020-01-28 06:55:13
解答
∫xsin^2xdx
=1/2∫x(1-cos2x)dx
=1/4x^2-1/2∫xcos2xdx
=1/4x^2-1/4∫xdsin2x
=1/4x^2-1/4xsin2x+1/4∫sin2xdx
=1/4x^2-1/4xsin2x-1/8cos2x+C
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