∫01dy∫0y2ycos(1一x)2dx=________.

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问题01dy∫0y2ycos(1一x)2dx=________.

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答案[*]sin1

解析01dy∫0y2ycos(1-x)2dx=∫01cos(1-x)2dx∫1ydy
=01(1一x)cos(1一x)2dx=01cos(1一x)2d(1一x)2
=
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