求解这道定积分 谢谢

2025-05-18 18:49:22
推荐回答(2个)
回答1:

设x=sint,t范围是[π/6,π/2]
原式=∫(π/6,π/2)csctdt=lnltan(t/2)l(π/6,π/2)
=ln1-lntan(π/12)
=lncot(π/12)
=ln(2-√3)

回答2:

令x=sint,则dx=costdt
原式=∫(π/6,π/2) costdt/sintcost
=∫(π/6,π/2) csctdt
=ln|csct-cott||(π/6,π/2)
=ln|1-0|-ln|2-√3|
=-ln(2-√3)
=ln(2+√3)