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求定积分∫(0到π/2)(asinx+bcosx)dx
人气:231 ℃ 时间:2020-04-09 15:46:00
解答
∫(0到π/2)(asinx+bcosx) dx
= ∫(0→π/2) (asinx + bcosx) dx
= -a * cosx + b *sinx (x = 0→π/2)
= a + b
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