∫1/(2+cos x) dx 定积分?

∫1/(2+cos x) dx 定积分?
∫1/(2+cos x) dx 定积分?

∫1/(2+cos x) dx 定积分?
设t=tan(x/2)
则cosx=[cos²(x/2)-sin²(x/2)]/[cos²(x/2)+sin²(x/2)]
=[1-tan²(x/2)]/[1+tan²(x/2)]
=(1-t²)/(1+t²)
dx=d(2arctant)=2dt/(1+t²)
故∫1/(2+cosx)dx=∫1/[2+(1-t²)/(1+t²)]*[2dt/(1+t²)]
=∫2dt/(3+t²)
=2/√3∫d(t/√3)/[1+(t/√3)²]
=2/√3arctan(t/√3)+C

把思路告诉你， 设u=tan(x/2) cosx 就等于 (1-u^)/(1+u^2) dx=d(2arctanu) 代进去可以化出来