看板 trans_math 關於我們 聯絡資訊
※ 引述《n0204159 (黃黑人)》之銘言: : ∫xtan^2x dx 從0積到π/4 : 請各位高手為不才解惑 : 謝謝!! 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 有錯請指正 <(_._)> -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 140.119.129.141
OKOK98:完美的解答~ 218.171.141.82 12/13 17:34
n0204159:請問第三個等式如何成為第四個等式.. 140.120.226.94 12/13 21:10
sexysam:分部積分 udv=uv-vdu 140.112.241.48 12/13 21:39
n0204159:嗯嗯! 感謝各位高手! 140.120.226.94 12/14 00:11