Inverse of $f(x)= x+\sin(x)$? How to find the inverse of $f(x) = x+\sin(x)$, analytically?

Well how should I proceed to find the inverse of $f(x)$? Basically I have applied graphical approach to solve the equation, but I want to know the inverse equation by analytical method.
Let $y = x + \sin x = f(x)$;
So, $x = f^{-1}(y)$. 
Differentiating the first expression according to $y$ leads to $x' + x'\cos(x) = 1$, or $x'(1+\cos x) = 1$.
Let us put $x=\arccos t$, so $x' = -t'/\sqrt{1-t^2}$. 
This leads to $-t'(1+t)/\sqrt{1-t^2} = 1$, or $t'=-\sqrt{\frac{1-t}{1+t}}$.
I believe that this is a standard differential equation. I don't know if it is solvable. If it is not, then the problem has no analytical closed expression.

Another way:
Using the formula $\cos x =  \frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}$, we are lead to the differential equation $x' = \frac{1+\tan^2(x/2)}{2}$. We can put $u = \tan(x/2)$, so $u'=\frac{1}{2}x'(1+u^2)$, and $x' = \frac{2u'}{1+u^2}$. So, the equation becomes $u' = \frac{(1+u^2)^2}{4}$. If I'm not wrong, there is a method to handle somehow this kind of equations. I will try tomorrow.
Let $y = x + \sin x = f(x)$;
So, $x = f^{-1}(y)$. 
Differentiating the first expression according to $y$ leads to $x' + x'\cos(x) = 1$, or $x'(1+\cos x) = 1$.
Let us put $x=\arccos t$, so $x' = -t'/\sqrt{1-t^2}$. 
This leads to $-t'(1+t)/\sqrt{1-t^2} = 1$, or $t'=-\sqrt{\frac{1-t}{1+t}}$.
I believe that this is a standard differential equation. I don't know if it is solvable. If it is not, then the problem has no analytical closed expression.

Another way:
Using the formula $\cos x =  \frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}$, we are lead to the differential equation $x' = \frac{1+\tan^2(x/2)}{2}$. We can put $u = \tan(x/2)$, so $u'=\frac{1}{2}x'(1+u^2)$, and $x' = \frac{2u'}{1+u^2}$. So, the equation becomes $u' = \frac{(1+u^2)^2}{4}$. If I'm not wrong, there is a method to handle somehow this kind of equations. I will try tomorrow.