求定积分∫_0^2π_ (√2a^2(1-cost))dt

题目

求定积分∫_0^2π_ (√2a^2(1-cost))dt
最好说明用了什么公式定理

答案

∫(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