求定积分∫(-π⼀4→π⼀4)xdx⼀(1+sinx)

∫xdx/(1+sinx)
=∫x*d[-2/（1+tan（x/2）] 分布积分法
=-2x/[1+tan（x/2）]-∫[-2dx/（1+tanx/2）]
=-2x/[1+tan（x/2）]+2∫dx/[1+tan（x/2）]
令u=x/2，du=1/2dx
=-2x/[1+tan（x/2）]+4∫du/（1+tanu）
令t=tanu，dt=sec^2udu=(1+tan^2u)du=(1+t^2)du
=-2x/[1+tan(x/2)]+4∫dt/(1+t)(1+t^2)
=-2x/[1+tan(x/2)]+4∫{[1/[2(1+t^2)-t/[2(1+t^2)]+1/[2(1+t)]}dt 待定系数法化简有理多项式
=-2x/[1+tan(x/2)]+2∫dt/(1+t^2)-2∫tdt/(1+t^2)+∫2dt/(1+t)
=-2x/[1+tan(x/2)]+2∫dt/(1+t^2)-∫d(1+t^2)/(1+t^2)+2∫d(1+t)/(1+t)
=-2x/[1+tan(x/2)]+2arctant-ln|1+t^2|+2ln|1+t|+C
=-2x/[1+tan(x/2)]+2u-ln|1+tan^2u|+2ln|1+tanu|+C
=-2x/[1+tan(x/2)]+x-2ln|sec(x/2)|+2ln|1+tan(x/2)|+C
=-2x/[1+tan(x/2)]+x+2ln|cos(x/2)[1+tan(x/2)]|+C
所以∫<-π/4,π/4>xdx/(1+sinx)
=-[2*π/4/(1+tanπ/8)-2*（-π/4)/[1+tan(-π/8)]-2[ln|cosπ/8*(1+tanπ/8)|-ln|cos(-π/8)*(1+tan-π/8)|]
=-π/2[1/（1+tan(π/8)+1/(1-tanπ/8)]-2ln|(1+tanπ/8)/(1-tanπ/8)|
带入tanπ/8=√2-1,结果为
π/2-2ln(√2-1)
在提供一种巧算：
=∫<-π/4,0>xdx/(1+sinx)+∫<0,π/4>xdx/(1+sinx)
对前部分设x=-t有dx=-dt
=∫<π/4,0>-t(-dt)/(1-sint)+∫<0,π/4>xdx/(1+sinx)
=-∫<0,π/4>tdt/(1-sint)+∫<0,π/4>xdx/(1+sinx)
=-∫<0,π/4>xdx/(1-sinx)+∫<0,π/4>xdx/(1+sinx)
=∫<0,π/4>[x/(1+sinx)-x/(1-sinx)]dx
=2∫<0,π/4>tanxsecx*xdx
=2∫<0,π/4>xd(secx)
=2xsecx-2∫<0,π/4>secxdx
=2xsecx-2ln|tan（π/4+x/2)|
=2*(π/4)*√2-2ln(tan3π/8-tanπ/4)
=(√2/2)π-2ln(√2-1)前面计算复杂最后一步结果因该有错。

∫（-a ,a）f(x)dx=∫(0.a)[f(x)+f(-x)] dx (书上定理)

∫(-π/4→π/4)xdx/(1+sinx)

=∫(0→π/4)x/(1+sinx) dx -∫(0→π/4)x/(1-sinx)dx (通分)
=∫(0→π/4)-2xsinx /(cosx)^2 dx
=∫(0→π/4)-2x d（1/cosx）
=-2x/cosx I(0→π/4) +2∫(0→π/4)1/cosxdx
=-√2 π/2 +2lnIsecx+tanxI I(0→π/4)
=-√2 π/2 +2ln(√2+1)
楼上的过于复杂，非常感谢楼上的回答