Question

求(x+sinx)/(1+cosx)在x(0,派/2)上的定积分,给个思路

Answer 1

答案为π/2,先做个不定积分然后代入上下限值∫((x+sinx)/(1+cosx))dx =∫[(x+sinx)/2cos?(x/2)]dx =∫(x+sinx)d(tan(x/2)) =(x+sinx)*tan(x/2)-∫tan(x/2)d(x+sinx)→分部积分法=xtan(x/2)+sinx*tan(x/2)-∫tan(x/2)(1+cosx)dx =xtan(x/2)+2sin(x/2)cos(x/2)*tan(x/2)-∫tan(x/2)*2cos?(x/2)dx =xtan(x/2)+2sin?(x/2)-∫sin(x)dx =xtan(x/2)+2sin?(x/2)-cos(x)+C =xtan(x/2)+K ∴∫(0到π/2) (x+sinx)/(1+cosx) dx=xtan(x/2) (0到π/2)=π/2*tan(π/2*1/2)-0=π/2*tan(π/4)=π/2