From the first Cauchy-Riemann condition we have:
$\frac{\partial v}{\partial y} = \frac{\partial u}{\partial x} = \sinh x \cos y$
$v(x,y) = \int \sinh x \cos y
\mathrm{d}y = \sinh x \sin y + F(x)$
$\frac{\partial v}{\partial x} = \cosh x \sin y + F'(x)$
$-\frac{\partial u}{\partial y} = \cosh x \sin y.$
Since the second Cauchy-Riemann condition requires:
$\frac{\partial v}{\partial x} = - \frac{\partial u}{\partial y},$
we have:
$F'(x) = 0,$
$F(x) = constant.$
The required function is then:
$v(x,y) = \sinh x \sin y + C.$
You did not compute the line integral correctly.
For $u(x,y)=e^{-y}\sin x$, we have $u_{,1}(x,y)=\dfrac{\partial u(x,y)}{\partial x}=e^{-y}\cos x$ and $u_{,2}(x,y)=\dfrac{\partial u(x,y)}{\partial y}=-e^{-y}\sin x$. Hence taking the straight-line path $\gamma\colon\tau\in[0,1]\mapsto(x\tau,y\tau)$ joining $(0,0)$ to $(x,y)$, we have
\begin{align*}
&\int_{(0,0)}^{(x,y)} (-u_{,2},u_{,1})(s, t)\cdot(\mathrm{d}s,\mathrm{d}t)\\
&=\int_0^1 (-u_{,2},u_{,1})(\gamma(\tau))\cdot\dot\gamma(\tau)\,\mathrm{d}\tau\\
&=\int_0^1 (-u_{,2},u_{,1})(x\tau,y\tau)\cdot\dot(x,y)\,\mathrm{d}\tau\\
&=\int_0^1 (e^{-y\tau}\sin (x\tau),e^{-y\tau}\cos x\tau)\cdot\dot(x,y)\,\mathrm{d}\tau\\
&=\int_0^1 (xe^{-y\tau}\sin (x\tau)+ye^{-y\tau}\cos x\tau)\,\mathrm{d}\tau\\
&=\Big[-e^{-y\tau}\cos(x\tau)\Big]_{\tau=0}^{\tau=1}=1-e^{-y}\cos x.
\end{align*}
Best Answer
It is easy to show that the Laplace equations is satisfied, $\nabla^2u=0$. Therefore, we can go about find the harmonic conjugate. By the Cauchy-Reimann equations, we have \begin{alignat}{2} u_x &=\cosh(x)\sin(y) &&{}= v_y\\ u_y &=\sinh(x)\cos(y) &&{}=-v_x \end{alignat} We can integrate $u_x = v_y$ by $y$ so $$ v(x,y) = \int\cosh(x)\sin(y)dy = -\cosh(x)\cos(y) + g(x)\tag{1} $$ Now, $-v_x = u_y$ so $$ -v_x = \sinh(x)\cos(y) - g'(x) = \sinh(x)\cos(y)\tag{2} $$ Can you take it from here? Next, you will be solving for $g(x)$ in equation $(2)$ and then plugging it into equation $(1)$ to identify the harmonic conjugate $v(x,y)$.