Question

请问这个函数如何积分？

（1/(cos(x) + a) ）dx 从0 到 2π 积分，谢谢！

Answer 1

∫(1/cosx + a)dx
=∫(1/cosx)dx + a*(2π - 0)
=2πa + ∫{1+[tan(x/2)]^2}/{1-[tan(x/2)]^2} dx 注：cosx = {1-[tan(x/2)]^2}/{1+[tan(2/x)]^2}
=2πa + 1/2*∫[sec(2/x)]^2 d(x/2)/{1-[tan(x/2)]^2} 注：(secα)^2 = 1 + (tanα)^2
=2πa + 1/2*∫d[tan(x/2)]/{1-[tan(x/2)]^2} 注：(tanα)' =(secα)^2
=2πa + 1/2*1/2*∫{1/[1-tan(x/2)] + 1/[1+tan(x/2)]} d[tan(x/2)] 注：1/(1-a^2) = 1/2*[1/(1-a) + 1/(1+a)]
=2πa + 1/4 * ln[1+tan(x/2)]/[1-tan(x/2)] |0~2π
=2πa + 1/4*(ln1 - ln1)
=2πa