Question

1/(1 cost)的不定积分怎么求

1/(1 cost)的不定积分怎么求

Answer 1

是∫ 1/(1+cost) dt?
=∫ (1-cost)/[(1+cost)(1-cost)] dt
=∫ (1-cost)/sin²t dt
=∫csc²t dt-∫csctcott dt
=-cott-(-csct)+C
=csct-cott+C
应该是这样的，我这样写的对了哦！

Answer 2

是∫ 1/(1+cost) dt?
=∫ (1-cost)/[(1+cost)(1-cost)] dt
=∫ (1-cost)/sin²t dt
=∫csc²t dt-∫csctcott dt
=-cott-(-csct)+C
=csct-cott+C

Answer 3

∫ 1/(1+cost) dt的话
先把cost变成2cos^(t/2)-1
然后就成了∫ 1/cos^(t/2) d(t/2)=∫sec^2 (t/2)d(t/2)=查积分表