Re: [考古] 96中山電機
※ 引述《acecaz ()》之銘言：
: 今天中山還不錯 電資大樓好像有開空調 還滿舒適的
: 趁亂超了一些不會做的題目 有高手能提示一下嗎~
: π/2 dx
: 6. Evaluate ∫ --------------
: 0 1+(tanx)^√2
π/2 1
∫ ------------------ dx
0 1 + (tanx)^(√2)
π/2 1
= ∫ --------------------------- dx
0 1 + ((sinx)/(cosx))^(√2)
π/2 (cosx)^(√2)
= ∫ ----------------------------- dx
0 (cosx)^(√2) + (sinx)^(√2)
π/2 (cosx)^(√2)
= ∫ ----------------------------- dx
0 (sinx)^(√2) + (cosx)^(√2)
π/2 (cosx)^(√2)
令 I = ∫ ----------------------------- dx
0 (sinx)^(√2) + (cosx)^(√2)
π/2 (cosx)^(√2)
則 I = ∫ ----------------------------- dx
0 (sinx)^(√2) + (cosx)^(√2)
0 (cos((π/2) - y))^(√2)
= ∫ (---------------------------------------------------)(-1) dy
π/2 (sin((π/2) - y))^(√2) + (cos((π/2) - y))^(√2)
(令 y = (π/2) - x , 則 x = (π/2) - y => dx = (-1) dy)
(x = 0 => y = π/2 , x = π/2 => y = 0)
π/2 (siny)^(√2)
= ∫ ----------------------------- dy
0 (cosy)^(√2) + (siny)^(√2)
π/2 (siny)^(√2)
= ∫ ----------------------------- dy
0 (siny)^(√2) + (cosy)^(√2)
π/2 (sinx)^(√2)
= ∫ ------------------------------ dx
0 (sinx)^(√2) + (cosx)^(√2)
π/2 (cosx)^(√2)
2I = ∫ --------------------------- dx
0 (sinx)^(√2) + (cosx)^(√2)
π/2 (sinx)^(√2)
+ ∫ --------------------------- dx
0 (sinx)^(√2) + (cosx)^(√2)
π/2 (cosx)^(√2) (sinx)^(√2)
= ∫ --------------------------- + --------------------------- dx
0 (sinx)^(√2) + (cosx)^(√2) (sinx)^(√2) + (cosx)^(√2)
π/2 (sinx)^(√2) + (cosx)^(√2)
= ∫ ----------------------------- dx
0 (sinx)^(√2) + (cosx)^(√2)
π/2 |π/2 π
= ∫ 1 dx = x | = ---
0 |0 2
π/2 (cosx)^(√2) π
I = ∫ --------------------------- dx = ---
0 (sinx)^(√2) + (cosx)^(√2) 4
π/2 1 π
∫ ------------------ dx = ---
0 1 + (tanx)^(√2) 4
: 1 (a^x)-1 1
: 8. Calculate lim (--- ---------)^(---) , where a>0 , a\=1
: x→∞ x a-1 x
: 9. 1^2 2^2 3^2 4^2 5^2
: ----- + ----- + ----- + ----- + ----- + ...... =
: 0! 1! 2! 3! 4!
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