Integral with hyperbolic functions I need to compute: 
$$ \int_{x^2+y^2=1} \frac{\sinh(x)dy- \sin(y)dx}{\cosh(x)-\cos(y)}$$ where the circle $x^2 + y^2 = 1$ is oriented anticlockwise. So, can somebody show me how? I found the antiderivative (which is equal to $2\arctan [\coth(\frac{x}{2}) \cdot \tan(\frac{y}{2})])$, but it's not defined for line $x=0 $ and line $y=0$. So I guess we need some tricky substitution here, but I have no idea. Please, may somebody help?
 A: But you are almost there! 
Take your antiderivative
$$ F(x,y)=2\arctan\left(\coth\frac{x}{2}\cdot\tan\frac{y}{2}\right)$$
and compute its value at some point of the unit circle. For example, let us take the point $A_+=(0^+,-1)$. The argument $g(x,y)=\coth\frac{x}{2}\cdot\tan\frac{y}{2}$ then tends to $-\infty$, we can fix $F(A_+)$ to be e.g. $-2\cdot\frac{\pi}{2}$ and define $\arctan$ as function with values in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. 
Next move counterclockwise requiring that $F(x,y)$ varies continuously. Note that since $y\in[-1,1]$, $\tan\frac{y}{2}$ is never infinite and we have to care only about $\coth\frac{x}{2}$. We will not encounter any problems, however, until we reach the point $B_+=(0^+,1)$ where $g\rightarrow +\infty$ and $F(B_+)=2\cdot\frac{\pi}{2}$. Hence the integral over the corresponding half of unit circle is $F(B_+)-F(A_+)=2\pi$.
Now pass from $B_+$ to $B_-=(0^-,1)$. Then $g(x,y)$ jumps from $+\infty$ to $-\infty$. Since we want to keep the same value of $F$ (equal to $\pi$), we now have to consider another branch of $\arctan$, namely taking the values in $\left(\frac{\pi}{2},\frac{3\pi}{2}\right)$. Continuing to move counterclockwise, we finally arrive to the point $A_-=(0^-,-1)$, where $g\rightarrow+\infty$ so that $F(A_-)=3\pi$. Hence the second half of the unit circle contributes $F(A_-)-F(B_-)=2\pi$ to the integral so that its total value is $4\pi$.
A: Given the integral:
$$\int_0^{2\pi} f(\theta) d\theta = \int_0^{2\pi} \frac{\cos(\theta) \sinh(\cos(\theta))+ \sin(\theta)\sin(\sin(\theta))}{\cosh(\cos(\theta))-\cos(\sin(\theta))} d\theta$$
Looking at $f(\theta-\pi/2) +f(\theta)$ you may see it's bounded (numerically) by $4 \pm \sim 0.004$. Thus, $f$ may be approximated by $f(x) \simeq 2 + g(x)$, where $g(x)=-g(x-\pi/2)$, and therefore the entire integral may be approximated by $4\pi$.
