求不定积分(cos2x)^2 dx
人气：471 ℃ 时间：2019-12-06 09:29:20
解答
∫ cos²2x dx
= (1/2)∫ (1 + cos4x) dx
= (1/2)[x + (1/4)sin4x] + C
= x/2 + (1/8)sin4x + C
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