Re: [問題] an integral question

[wuling510665 (摩卡金幣)]

※ 引述《i563214f (i563214f)》之銘言：
∫ ln(sin(x))dx x from 0 to π/2

please tell me how to calculate this question

thank you

Let  I=∫ln(sinx)dx x from 0 to pi/2    ---1
PROPERTY:∫f(x)dx    x from 0 to a
        =∫f(a-x)dx  x from 0 to a
     I=∫ln(sin(pi/2-x))dx=∫ln(cosx)dx ---2

1+2:2I=∫(lnsinx+lncosx)dx
      =∫ln(sinxcosx)dx
      =∫ln(sin2x/2)dx
      =∫(ln(sin2x)-ln2)dx
      =∫(ln(sin2x)dx-∫(ln2)dx
      =∫(ln(sin2x)dx-ln2*x|x from 0 to pi/2
      =1/2∫ln(sinu)du-ln2*pi/2  u from 0 to pi
      =1/2*2∫ln(sinu)du-ln2*pi/2 (PROPERTY)
      =∫ln(sinu)du-ln2*pi/2
      =I-ln2*pi/2

    2I=I-ln2*pi/2
     I=ln2*(-pi/2)
以上= =


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◆ From: 140.112.211.173
→ wuling510665:倒數第六行的x from 0 to pi/2
→ tp61i6e04   :ln(x)  x不能等於0 這篇是....
→ wuling510665:x沒有等於0啊!只是從0積到pi/2
→ tp61i6e04   :從0積到pi/2不用等於0= =?
→ tp61i6e04   :他根本不會形成封閉的區塊...哪來的面積
→ tp61i6e04   :好ㄅ~我錯了
推 waanapple   :這是有名難題~如果有人能不看解答解出來~真的超強!!