求解一题高数定积分

Question

如题所述

Answer 1

令t=2x x=t/2 dx=dt/2

原式=(1/2)*∫(0,8π/3) dt/(2+cost)
令u=tan(t/2) cost=(1-u^2)/(1+u^2) dt=2du/(1+u^2)
原式=∫(0,√3) du/(2+2u^2+1-u^2)
=∫(0,√3) du/(3+u^2)
=(1/√3)*arctan(u/√3) |(0,√3)
=(√3/12)π