∫(0,π/2)ln[(x+sinx)/(1+cosx)]dxT_T

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∫(0,π/2)ln[(x+sinx)/(1+cosx)]dxT_T

∫(0,π/2)ln[(x+sinx)/(1+cosx)]dxT_T
∫(0,π/2)ln[(x+sinx)/(1+cosx)]dx
T_T

∫(0,π/2)ln[(x+sinx)/(1+cosx)]dxT_T
令f(x)=x+sinx,则f'(x)=1+cosx.
再令F(x)=∫f'(x)/f(x) dx =∫1/f(x) df(x)=ln(f(x)).
而f'(x)/f(x)=F'(x).
ln[(x+sinx)/(1+cosx)]
= -ln[(1+cosx)/(x+sinx)]
= -ln[F'(x)].

∫(0,π/2)ln[(x+sinx)/(1+cosx)]dx
= -∫ln[F'(x)]dx
= -x·ln[F'(x)]+∫x·F''(x)/F'(x) dx