Question

求定积分问题

这个函数怎么求他的原函数？

Answer 1

∫(-π/4->π/4) 4xtanx/(cosx)^2 dx

=2∫(0->π/4) 4xtanx/(cosx)^2 dx
=2∫(0->π/4) 4x.sinx/(cosx)^3 dx
=4∫(0->π/4) x d [1/(cosx)^2]
= 4[x/(cosx)^2]|(0->π/4) -4∫(0->π/4) (secx)^2 dx
=2π -4[tanx]|(0->π/4)
=2π -4

追问

谢谢

Answer 2

∫4xtanx /(cosx^2) dx

=∫4x sinx/(cosx)^3 dx
=-∫4x/(cosx)^3 dcosx
=∫2xd(1/(cosx)^2
=2x/(cosx)^2 - 2∫(secx)^2 dx
=2x/(cosx)^2 - 2tanx +C

追问

谢谢