题目

定积分∫(x/(1+sin(x)))dx 从-π/4积到π/4
要过程或讲下思路

提问时间：2020-10-31

答案

∫(- π/4→π/4) x/(1 + sinx) dx
= ∫(- π/4→π/4) x(1 - sinx)/cos²x dx
= ∫(- π/4→π/4) (x - xsinx)/cos²x dx
= ∫(- π/4→π/4) xsec²x dx - ∫(- π/4→π/4) xsecxtanx dx
= ∫(- π/4→π/4) x d(tanx) - ∫(- π/4→π/4) x d(secx)
= xtanx - ∫(- π/4→π/4) tanx dx - xsecx + ∫(- π/4→π/4) secx dx
= [(π/4) - (- π/4)(- 1)] + lncosx - [(π/4)√2 - (- π/4)√2] + ln(secx + tanx)
= ln(√2/2) - ln(√2/2) - π/√2 + ln(√2 + 1) - ln(√2 - 1)
= ln[(√2 + 1)/(√2 - 1)] - π/√2
= ln(3 + 2√2) - π/√2
过程先上下乘以1 - sinx
然后分裂为两个积分,之后它们分别运用分部积分法
就这么简单