求1除以(2+5cosx)的不定积分用万能代换

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)]}

Answer 2

∫1/(2+5cosx)dx=∫1/(10cos²(x/2)-3)dx=∫10sec²(x/2)/(10-3sec²(x/2))dx=∫20/(7-3tan²(x/2))dtan(x/2)
令t=tan(x/2)，=20/7∫1/(1-(t√(3/7))²)dt=20/7√(7/3)∫1/(1-(t√(3/7))²)dt√(3/7)，令m=t√(3/7)，=20/√21∫1/(1-m²)dm=10/√21∫1/(1+m)+1/(1-m)dm=(10/√21)ln|(m+1)/(m-1)|+C

Answer 3

(10√21)ln|[(√3)tan(x/2)+√7]/[(√3)tan(x/2)-√7]|/√21+c~