What we learned in complex analysis is
$$\int_{\gamma}f(z)\,dz=\int_{\gamma}(u(x,y)+iv(x,y))(dx+i\,dy)$$
if we let $f(z)=u+iv$ and $dz=dx+i\,dy$.
So how could I calculate $f(z)=\sin(z)$ ? What is the corresponding part of $u$ and $v$ for $f(z)=\sin(z)$?
Your problem is to find $$\Re(\sin(z))\quad\text{and}\quad\Im(\sin(z))$$
Plug $z=x+yi$ to get $$\sin(x+yi)=\sin(x)\cos(yi)+\cos(x)\sin(yi)$$
Now use the formulas $\sin(yi)=i\sinh(y)$ and $\cos(yi)=\cosh(y)$ to get
$$\sin(x+yi)=\underbrace{\sin(x)\cosh(y)}_\text{real part}+i\underbrace{\cos(x)\sinh(y)}_\text{imaginary part}$$
So your corresponding functions are equal to $$u(x,y)=\sin(x)\cosh(y)$$ $$v(x,y)=\cos(x)\sinh(y)$$