Re: [微積] 請教一下兩題

[mack (腦海裡依然記得妳)]

※ 引述《chribaby (過去˙現在˙未來)》之銘言：
: http://www.lib.ntu.edu.tw/exam/graduate/91/91049.htm
: 2. 設f(x)=x^nln(x)，則f^(n+1)(x)=?
f^(1)(x) = nx^(n-1)ln(x)+x^n*(1/x) = nx^(n-1)ln(x)+x^(n-1)
         = n*[ln(x)+(1/n)]x^(n-1)
f^(2)(x) = n(n-1)*[ln(x)+(1/n)]x^(n-2)+nx^(n-2)
         = n(n-1)*[ln(x)+(1/n) + 1/(n-1)]x^(n-2)
f^(3)(x) = n(n-1)(n-2)*[ln(x)+(1/n) + 1/(n-1)]x^(n-3)+n(n-1)x^(n-3)
         = n(n-1)(n-2)*[ln(x) + 1/n + 1/(n-1) + 1/(n-2)]x^(n-3)
         .
         .
         .
f^(n)(x) = (n!/1!)*[ln(x) + 1/n + 1/(n-1) +...+ 1/2]x
f^(n+1)(x) = (n!)*[ln(x) + 1/n + 1/(n-1) +...+ 1/2]+(n!/1!)
           = (n!)*[ln(x) + 1/n + 1/(n-1) +...+ 1/1]
: 這題感覺要用泰勒(是這樣嗎?)
: 但卻不知道要怎麼下手
: 拜託幫幫我~~
: 4. 定積分 S(pi/4~pi/2)1/(sinx)^4 dx
 π/2                 π/2
∫   1/(sinx)^4 dx = ∫   (cscx)^2*(cscx)^2 dx
 π/4                 π/4
                       π/2
                   = -∫   (cscx)^2 d(cotx)
                       π/4
                       π/2
                   = -∫   [1+(cotx)^2] d(cotx)
                       π/4
                                         |π/2
                   = -[(cotx)+(cotx)^3/3]|π/4
                   = -[0+0/3]+[1+1/3]
                   = 4/3


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