求定积分2π∫_0^2π_ (√2)a^2(1-cost)^(3/2) dt

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/21 14:24:19

求定积分2π∫_0^2π_ (√2)a^2(1-cost)^(3/2) dt
要过程啊,最好说明用了什么公式
答案是(64/3)πa^2

2π∫(0→2π) √2a²(1 - cost)^(3/2) dt
= 2√2πa²∫(0→2π) √2a²[2sin²(t/2)]^(3/2) dt,cosx = 1 - 2sin²(x/2) => 1 - cosx = 2sin²(x/2)
= 2√2πa²∫(0→2π) 2^(3/2) * |sin³(t/2)| dt
= 16πa²∫(0→2π) sin³(t/2) d(t/2),在t∈[0,2π],sin³(t/2) ≥ 0,∴|sin³(t/2)| = sin³(t/2)
= 16πa²∫(0→2π) - [1 - cos²(t/2)] d[cos(t/2)],∫ sinx dx = - cosx + C
= 16πa²[- cos(t/2) + (1/3)cos³(t/2)] |(0→2π)
= 16πa²[(- (- 1) + (1/3)(- 1)) - (- 1 + (1/3)(1))],cos(π) = - 1,cos(0) = 1
= (16πa²)(4/3)
= (64/3)πa²

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