求定积分[-π/3,π/3]∫[(3/4)tan²u+(3√3/2)tanu+9/4]du
原式=[-π/3,π/3]∫[(3/4)tan²u+9/4]du+[-π/3,π/3]∫[(3√3/2)tanu]du
=[-π/3,π/3]∫[(3/4)tan²u+9/4]du+0 (tanu是奇函数,在对称区间上的积分=0)
=[-π/3,π/3]∫(3/4)(tan²u+1)du+[-π/3,π/3]∫(6/4)du
=[-π/3,π/3](3/4)∫(sec²udu+[-π/3,π/3]∫(3/2)du
=[(3/4)tanu+(3/2)u]︱[-π/3,π/3]=(3/4)[tan(π/3)-tan(-π/3)]+(3/2)[(π/3)-(-π/3)]
=(3/4)(√3+√3)+(3/2)(π/3+π/3)=[(3/2)√3]+π