∫(π/3 to π/6) (sint*cott/sect) dt?
回答 (1)
2016-03-09 4:02 pm
✔ 最佳答案
cott=cost/sint ; sect=1/cost∴ sint*cott/sect=cos^2t
cos^2t=1/2(1+cos2t)
∫(π/3 to π/6) (sint*cott/sect) dt = 1/2∫(π/3 to π/6) (1+cos2t) dt
=1/2[t+1/2sin2t](π/3 to π/6)
=1/2[(π/6+sqrt(3)/4)-(π/3+sqrt(3)/4)]
=-π/12
收錄日期: 2021-04-30 20:10:28
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160309045423AAdQJrG