求定积分∫(-π/4→π/4)xdx/(1+sinx)
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求定积分∫(-π/4→π/4)xdx/(1+sinx)
∫[-π/4,π/4]x/(1+sinx) dx
= ∫[-π/4,π/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√2
= ∫[-π/4,π/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√2
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