1樓：匿名使用者
|∫[-π
/4，π/4]x/(1+sinx) dx
= ∫[-π/4，π版/4]x(1-sinx)/(1-sin²x) dx
= ∫[-π/4，π/4]x(1-sinx)/cos²x dx
= ∫[-π/4，π/4]xsec²x dx - ∫[-π/4，π/4]xsecxtanx dx
= ∫[-π/4，π/4]x dtanx - ∫[-π/4，π/4]x dsecx
= xtanx - ∫[-π/4，π/4]tanx dx - xsecx + ∫[-π/4，π/4]secx dx
= [π/4 - π/4] + cosx - [π/4*√2 + π/(2√2)] + ln|secx+tanx|
= [1/√2 - 1/√2] - π/√2 + ln(sec(π/4)+tan(π/4) - ln(sec(-π/4)+tan(-π/4))
= -π/√2 + ln(√2+1) - ln(√2-1)
= ln(3+2√2) - π√權2
計算定積分∫(0→π)(xsinx)/(1+sin^2x)dx
2樓：匿名使用者
令x = π - y、dx = - dy
x = 0 → y = π
x = π → y = 0
m = ∫[0→π
636f707962616964757a686964616f31333330363837] (xsinx)/(1 + sin²x) dx
= ∫[π→0] [(π - y)sin(π - y)]/[1 + sin²(π - y)] (- dy)
= ∫[0→π] [(π - x)sinx]/(1 + sin²x) dx
= π∫[0→π] sinx/(1 + sin²x) dx - m
2m = π∫[0→π] sinx/[1 + (1 - cos²x)] dx
m = (- π/2)∫[0→π] d(cosx)/(2 - cos²x)
= (π/2)[1/(2√2)]∫[0→π] [(cosx + √2) - (cosx - √2)]/[(cosx - √2)(cosx + √2)] d(cosx)
= [π/(4√2)]∫[0→π] [1/(cosx - √2) - 1/(cosx + √2)] d(cosx)
= [π/(4√2)]ln[(cosx - √2)/(cosx + √2)] |[0→π]
= [π/(4√2)]
= [π/(2√2)]ln[(√2 + 1)/(√2 - 1)] ≈ 1.9579
3樓：李榮華
給個郵箱吧，發手寫版的
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