Question

求解这道定积分 谢谢

Answer 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)

Answer 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)