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

显示全部楼层 · 2012-6-15 12:28:51

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) ...

千问 · 2012-6-15 12:28:51

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...