1/(2+cosx)的不定积分是什么

Question

1/(2+cosx)的不定积分是什么

Answer 1

∫dx/(2+cosx)
cosx+1=2cos^2(x/2)
原式=∫dx/(2cos^2(x/2)+1)
=∫sec^(x/2)dx/(2+sec^2(x/2))
=2∫d(tg^(x/2))/(3+tg^2(x/2))
设tgx/2=t
=2∫dt/(1+(t/根号3)^2)
=2根号3∫d(t/根号3)/(1+(t/根号3)^2)
令t/根号3=tga
2根号3∫d(t/根号3)/(1+(t/根号3)^2)
=2根号3∫d(tga)/(1+(tga)^2)
=2根号3∫sec^2a/(sec^2a)da
=2根号3*a+c
a=arctg(t/根号3)
tg(x/2)=t
a=arctan(tg(x/2)/根号3))
原式=2根号（3）/3*arctan{（根号（3）/[3tan(x/2)]}