考研数学多元函数微分题目

如图

(16a^4/3) ∫ (0->2π) [sin (t/2) ]^8 dt 表示不懂

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t/2 = u
dt = 2du
t=0, u=0
t=2π, u=π
(16a^4/3) ∫ (0->2π) [sin (t/2) ]^8 dt
=(16a^4/3) ∫ (0->π) (sinu)^8 .(2du)
=(32a^4/3) ∫ (0->π) (sinu)^8 du
=(32a^4/3) [∫ (0->π/2) (sinu)^8 du + ∫ (π/2->π) (sinu)^8 du ]
=(32a^4/3) [∫ (0->π/2) (sinu)^8 du + ∫ (0->π/2) (sinu)^8 du ]
=(64a^4/3) .∫ (0->π/2) (sinu)^8 du
let
y=π -u
dy = -du
u=π/2 , y=π/2
u=π , y=0
∫ (0->π/2) (sinu)^8 du ]
=∫ (π/2->0) (siny)^8 (-dy)
=∫ (0->π/2) (siny)^8 dy
=∫ (0->π/2) (sinu)^8 du