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求定积分∫_0^2π_ (√2a^2(1-cost))dt
最好说明用了什么公式定理
人气:199 ℃ 时间:2020-04-15 06:14:27
解答
∫(0~2π) √[2a²(1 - cost)] dt
= √2a∫(0~2π) √(1 - cost) dt
= √2a∫(0~2π) √[2sin²(t/2)] dt <== 定理:2sin²x = 1 - cos2x
= 2a∫(0~2π) |sin(t/2)| dt
= 2a∫(0~2π) sin(t/2) dt,当t∈[0,2π],sin(t/2) > 0
= 2a * -2cos(t/2) |_0^2π <== ∫ sinx dx = -cosx + C
= -4a * [cos(π) - cos(0)]
= -4a * (-1 - 1)
= 8a
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