求不定积分

2025-01-20 13:27:56
推荐回答(1个)
回答1:

设x=2sect
原式=∫sectdt
=∫1/costdt
=∫cost/(cost)^2dt
=∫1/[1-(sint)^2]dsint
=1/2∫1/(1+sint)-1/(sint-1)dt
=1/2ln|(1+sint)/(1-sint)|+C
=1/2ln|(1+sint)^2/(cost)^2|+C
=ln|(1+sint)/cost|+C
=ln|sect+tant|+C
=ln|x/2+√(x^2-4)/2|+C
=ln|x+√(x^2-4)|+C1
其中C1=C-ln2