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
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/01 22:36:35
![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](/uploads/image/z/5559892-52-2.jpg?t=%E6%B1%82%E5%AE%9A%E7%A7%AF%E5%88%862%CF%80%E2%88%AB_0%5E2%CF%80_+%28%E2%88%9A2%29a%5E2%281-cost%29%5E%283%2F2%29+dt%E8%A6%81%E8%BF%87%E7%A8%8B%E5%95%8A%2C%E6%9C%80%E5%A5%BD%E8%AF%B4%E6%98%8E%E7%94%A8%E4%BA%86%E4%BB%80%E4%B9%88%E5%85%AC%E5%BC%8F%E7%AD%94%E6%A1%88%E6%98%AF%2864%2F3%29%CF%80a%5E2)
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
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
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
0≤t≤2π,
∴0≤t/2≤π
∴sin﹙t/2﹚≥0
(1-cost)^(3/2) =【2sin²﹙t/2﹚】^(3/2) =﹙2√2﹚sin³﹙t/2﹚
2π∫_0^2π_ (√2)a^2(1-cost)^(3/2) dt
=2π∫_0^2π_ (√2)a²﹙2√2﹚sin³﹙t/2﹚dt
=2π∫_0^...
全部展开
0≤t≤2π,
∴0≤t/2≤π
∴sin﹙t/2﹚≥0
(1-cost)^(3/2) =【2sin²﹙t/2﹚】^(3/2) =﹙2√2﹚sin³﹙t/2﹚
2π∫_0^2π_ (√2)a^2(1-cost)^(3/2) dt
=2π∫_0^2π_ (√2)a²﹙2√2﹚sin³﹙t/2﹚dt
=2π∫_0^2π_ 4a²sin²﹙t/2﹚sin﹙t/2﹚dt
=2π∫_0^2π_ 4a²【1-cos²﹙t/2﹚】d【-2cos﹙t/2﹚】
=-16πa²∫_0^2π_ 【1-cos²﹙t/2﹚】d【cos﹙t/2﹚】
=-16πa² 【cos﹙t/2﹚-⅓ cos³﹙t/2﹚】|〈0,2π〉
= -16πa² 【(cosπ-⅓ cos³π﹚-﹙cos0-⅓ cos³0﹚】
= -16πa² 【(-1+⅓﹚-﹙1-⅓﹚】
= -16πa² ×(-4/3﹚
=(64/3)πa^2
收起