※ 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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