Question

求以下积分

∫x(sinx)^2n/{(sinx)^2n+(cosx)^2n} 积分区间(0,π) n∈N+

Answer 1

首先用一个公式：∫[0→π] xf(sinx) dx = (π/2)∫[0→π] f(sinx) dx
∫[0→π] x(sinx)^2n/[(sinx)^2n+(cosx)^2n] dx
=(π/2)∫[0→π] (sinx)^2n/[(sinx)^2n+(cosx)^2n] dx
=(π/2)∫[0→π/2] (sinx)^2n/[(sinx)^2n+(cosx)^2n] dx+(π/2)∫[π/2→π] (sinx)^2n/[(sinx)^2n+(cosx)^2n] dx
对后一项换元，令x=u+π/2，u：0→π/2
=(π/2)∫[0→π/2] (sinx)^2n/[(sinx)^2n+(cosx)^2n] dx
+(π/2)∫[0→π/2] (sin(u+π/2))^2n/[(sin(u+π/2))^2n+(cos(u+π/2))^2n] du
=(π/2)∫[0→π/2] (sinx)^2n/[(sinx)^2n+(cosx)^2n] dx
+(π/2)∫[0→π/2] (cosu)^2n/[(sinu)^2n+(cosu)^2n] du
将u写成x
=(π/2)∫[0→π/2] (sinx)^2n/[(sinx)^2n+(cosx)^2n] dx
+(π/2)∫[0→π/2] (cosx)^2n/[(sinx)^2n+(cosx)^2n] dx
=(π/2)∫[0→π/2] 1 dx
=π²/4