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/咐困拿(cosx)^2 dx
=2∫衡搭(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->尺肢π/4) (secx)^2 dx
=2π -4[tanx]|(0->π/4)
=2π -4

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谢谢

Answer 2

∫4xtanx /(cosx^2) dx

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

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谢谢