一道定积分题c

分部积分法
∫(0～π)f(x)dx
＝－∫(0～π)f(x)d(π－x)
＝－(π－π)f(π)＋(π－0)f(0)＋∫(0～π) (π－x)f'(x)dx
＝∫(0～π) (π－x)f'(x)dx
＝∫(0～π) (π－x)×sinx/(π－x)dx
＝∫(0～π) sinxdx
＝2

S[0到pi]f(x)dx

= S[0到pi]dx S[0到x] sintdt/(pi-t) [ 写成二重积分]

= S[0到pi]dt S[t到pi]sintdx/(pi-t) [ 交换积分顺序]

= S[0到pi] [sint/(pi-t)]dt S[t到pi]dx

= S[0到pi] [sint]dt

= 1