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求不定积分,xsin^2x
人气:473 ℃ 时间:2020-06-27 11:38:48
解答
∫x(sinx)^2dx
=(1/2)∫x(1-cos2x)dx
=(1/4)x^2-(1/4)∫xdsin2x
=(1/4)x^2-(1/4)(xsin2x)+(1/4)∫sin2xdx
=(1/4)x^2-(1/4)(xsin2x)+(-1/8)cos2x+C
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