Question

定积分 求此题详解过程

后面少打了个du

Answer 1

求定积分[-π/3，π/3]∫[(3/4)tan²u+(3√3/2)tanu+9/4]du
解：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

Answer 2

中间的3√3/2tanu是奇函数，积分区间对称，因此这个函数的积分为0.
原式=∫[-π/3，π/3] (3/4tan²u+9/4) du
=∫[-π/3，π/3] (3/4tan²u+3/4+6/4) du
=∫[-π/3，π/3] (3/4sec²u+3/2) du
=3/4tanu+3/2u [-π/3，π/3]
=3√3/4+(3/2)*(π/3)+3√3/4+(3/2)*(π/3)
=3√3/2+π