→ noelyang:太感激了!!! 218.166.216.10 03/27 12:14
※ 引述《noelyang (123)》之銘言：
: π/2 π/2
: ∫ ∫ siny/y dydx
: 0 x
Dirichlet 積分變換
π/2 π/2 π/2 y
∫ ∫ siny/y dydx = ∫ ∫ siny/y dxdy
0 x 0 0
π/2 siny x=y π/2
= ∫ x ---- | dy = ∫ siny dy = 1
0 y x=0 0
: 1
: ∫x^5(1-x^3)^1/3 dx
: 0
1
= (1/3)*∫x^3(1-x^3)^1/3 dx^3
0
let u = x^3
1 1
(1/3)*∫u (1-u )^1/3 du = (1/3)* ∫ [1-(1-u)](1-u )^1/3 du
0 0
1
= [ (1/7)*(1-u )^7/3 - (1/4)*(1-u )^4/3 ] | = 3/28
0
: 算半天都算不出來
: 感謝回答喔!!!
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