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∫xcos(x-1)dx不定积分
人气:337 ℃ 时间:2020-05-29 23:30:38
解答
∫xcos(x-1)dx=∫cos(x-1)d(x^2/2)=cos(x-1)*x^2/2-∫xd(cos(x-1)+c=cos(x-1)*x^2/2+∫sin(x-1)d(x^2/2)+c=cos(x-1)*x^2/2+∫sin(x-1)*x^2/2)+c=cos(x-1)*x^2/2+sin(x-1)*x^2/2-∫xcos(x-1)dx+c到此方程两边都有∫xc...
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