Question

求（x+sinx）/(1+cosx)在0到π/2的积分

Answer 1

法1
因为不定积分
∫(x+sinx)/(1+cosx)dx
=∫[x+2sin(x/2)cos(x/2)]/[2cos²(x/2)]dx
=∫[x/(2cos²(x/2))]dx+∫[2sin(x/2)cos(x/2)]/[2cos²(x/2)]dx
=∫xdtan(x/2)+∫tan(x/2)dx
=xtan(x/2)-∫tan(x/2)dx+∫tan(x/2)dx
=xtan(x/2)+C
所以原定积分
=xtan(x/2)|(0,π/2)
=π/2
法2请点击看大图哈