作者LuisSantos ( )
看板trans_math
标题Re: [考古] 97台大微B 第二题
时间Sat Dec 13 17:06:50 2008
※ 引述《n0204159 (黄黑人)》之铭言：
: ∫xtan^2x dx 从0积到π/4
: 请各位高手为不才解惑
: 谢谢!!
π/4 2
∫ (x)(tan x) dx
0
π/4 2
= ∫ (x)(sec x - 1) dx
0
π/4 2 π/4
= ∫ (x)(sec x) dx - ∫ x dx
0 0
π/4 x^2 |π/4
= ∫ (x) d(tanx) - ----- |
0 2 |0
|π/4 π/4 π^2
= (x)(tanx) | - ∫ tanx dx - ------
|0 0 32
π π/4 π^2
= --- - ∫ tanx dx - ------
4 0 32
π/4 π π^2
= -∫ tanx dx + --- - ------
0 4 32
|π/4 π π^2
= -ln|secx| | + --- - ------
|0 4 32
-1 π π^2
= (---)(ln2) + --- - ------
2 4 32
有错请指正
<(_._)>
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◆ From: 140.119.129.141
1F：推 OKOK98:完美的解答~ 218.171.141.82 12/13 17:34
2F：推 n0204159:请问第三个等式如何成为第四个等式.. 140.120.226.94 12/13 21:10
3F：推 sexysam:分部积分 udv=uv-vdu 140.112.241.48 12/13 21:39
4F：推 n0204159:嗯嗯! 感谢各位高手! 140.120.226.94 12/14 00:11