求x/1+sinx在0到π上的定积分

令x = π - u,dx = - du
x = 0,u = π
x = π,u = 0
N = ∫(0→π) x/(1 + sinx) dx
= ∫(π→0) (π - u)/[1 + sin(π - u)] * (- du)
= ∫(0→π) (π - u)/(1 + sinu) du
= ∫(0→π) (π - x)/(1 + sinx) dx
= π∫(0→π) dx/(1 + sinx) - N
2N = π∫(0→π) (1 - sinx)/[(1 + sinx)(1 - sinx)] dx
N = (π/2)∫(0→π) (1 - sinx)/cos²x dx
= (π/2)∫(0→π) (sec²x - secxtanx) dx
= (π/2)[tanx - secx]：(0→π)
= (π/2)[(0 - (- 1)) - (0 - (1))]
= π