作者LuisSantos (^______^)
看板trans_math
标题Re: [积分]
时间Tue Apr 24 11:34:18 2007
※ 引述《king911015 (早已放弃爱上你)》之铭言：
: 1 x
: Evaluate lim --- ∫ (1 - tan2t)^1/t dt
: x→0 X 0
1 x
lim (---)(∫ (1 - tan2t)^(1/t) dt)
x→0 x 0
x
∫ (1 - tan2t)^(1/t) dt
0
= lim ---------------------------
x→0 x
(1 - tan2x)^(1/x)
= lim --------------------
x→0 1
= lim (1 - tan2x)^(1/x)
x→0
= lim e^(ln((1 - tan2x)^(1/x)))
x→0
= e^(lim ln((1 - tan2x)^(1/x)))
x→0
ln(1 - tan2x)
= e^(lim ---------------)
x→0 x
((-2)((sec(2x))^2))/(1 - tan2x)
= e^(lim ---------------------------------)
x→0 1
(-2)((sec(2x))^2)
= e^(lim -------------------)
x→0 1 - tan2x
(-2)(1^2)
= e^(-----------)
1 - 0
-2
= e^(-----) = e^(-2)
1
--
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