高数定积分问题

=∫[0,π/2]xdx/[2cos²(x/2)] - ∫[0,π/2]d(cosx)/(1+cosx)
=∫[0,π/2]xd(x/2)/cos²(x/2) - ln(1+cosx)|[0,π/2]
=∫[0,π/2]x d[tan(x/2)] - (ln1-ln2)
=x tan(x/2)|[0,π/2] - ∫[0,π/2]tan(x/2) dx +ln2
=π/2 -0 +2∫[0,π/2] d[cos(x/2)]/cos(x/2) +ln2
=π/2 +2ln|cos(x/2)||[0,π/2]+ln2
=π/2 +2 [ln(√2/2) -ln1]+ln2
=π/2 + 2ln(√2/2)+ln2
=π/2 -2ln√2+ln2
=π/2

这个是一个例题（很多教材上都有，比如同济版高数）
∫(0～π）xf(sinx)dx＝π/2×∫(0～π）f(sinx)dx
推导的思路是左边积分换元：t＝π－x

∫(0->π/2) (x+sinx)/(1+cosx) dx
=∫(0->π/2) x/(1+cosx) dx + ∫(0->π/2) sinx/(1+cosx) dx

=∫(0->π/2) x/(1+cosx) dx - [ln|1+cosx|]|(0->π/2)

=∫(0->π/2) x/(1+cosx) dx + ln2
=(1/2)∫(0->π/2) x/[(1+cosx)/2] dx + ln2
=(1/2)∫(0->π/2) x . [sec(x/2)]^2 dx + ln2
=∫(0->π/2) x . dtan(x/2) + ln2
=[x.tan(x/2)]|(0->π/2) - ∫(0->π/2) tan(x/2) dx + ln2
=π/2 + 2[ln|cos(x/2)|]|(0->π/2) + ln2
=π/2 -ln2 + ln2
=π/2