Question

求定积分，积分上限2分之派，积分下限0，被积表达式[(x+sinx)/（1+cosx）]dx，答案是2分之派，怎么做

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