Question

求第1题怎么算？

Answer 1

然后把上下限带进去

原式=（tanπ/3-cotπ/3）-（tanπ/4-cotπ/4）

=√3-√3/3-0

=2√3/3

Answer 2

∫(π/4->π/3) dx/ (sinx.cosx)^2
=4∫(π/4->π/3) dx/ (sin2x)^2
=4∫(π/4->π/3) (csc2x)^2 dx
=-2[cot2x]|(π/4->π/3)
=-2 ( -√3/3 - 0)
=2√3/3