Finding harmonic conjugates by finding $du$ and $dv$ I want to solve the following problem:
For $u(x, y)=\sinh x \cos y$, find $d u$, find $d v$, and find $v$, the conjugate harmonic function of $u$.
I see the solution is something like this:
Use Cauchy-Riemanns equations page $83 .$
$$
\begin{aligned}
&u(x, y)=\sinh x \cos y \\
&d u=\cosh x \cos y d x-\sinh x \sin y d y \\
&d v=\sinh x \sin y d x+\cosh x \cos y d y \\
&v(x, y)=\cosh x \sin y
\end{aligned}
$$
My question is, how did the solution go from the $dv$ to $v$ ? You just integrate the the $dx$ and $dy?$ and add them together?  If so, the Integral of $\sinh x \sin y dx $ is $\sin \left(y\right)\cosh \left(x\right)$ and the integral of $\cosh x \cos y d y$ is $\cosh \left(x\right)\sin \left(y\right)$ ? If I add the two parts together don't I get that $2\cdot\cosh \left(x\right)\sin \left(y\right)$ ? Why is the answer just $\cosh \left(x\right)\sin \left(y\right)$?
 A: Cauchy–Riemann equations are
$$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}\tag{1}\label{1}$$
$$\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}\tag{2}\label{2}$$
We have $u=u(x,y)=\sinh(x)\cos(y)$, so
$$\frac{\partial u}{\partial x}=\cosh(x)\cos(y)\tag{3}\label{3}$$
and from $(1)$ we conclude that
$$\frac{\partial v}{\partial y}=\cosh(x)\cos(y)\tag{4}\label{4}$$
Now integrate both sides of (4) (this is actually partial integration - the opposite operation to partial differentiation, but this term isn't widely used for some reason):
$$v=\int\frac{\partial v}{\partial y}\,dy=\int\cosh(x)\cos(y)\,dy=\cosh(x)\sin(y)+C(x)\tag{5}\label{5}$$
The function $C(x)$ in $(5)$ is yet to be determined (it might depend on $x$, but doesn't depend on $y$). We'll find function $C(x)$ from $(2)$ and for this reason we need to find $\frac{\partial u}{\partial y}$ first:
$$\frac{\partial u}{\partial y}=-\sinh(x)\sin(y)\tag{6}\label{6}$$
From $(2)$ and $(6)$ we conclude that function $v$ has to satisfy this:
$$\frac{\partial v}{\partial x}=\sinh(x)\sin(y)\tag{7}\label{7}$$
Now let's put $(5)$ into $(7)$:
$$\frac{\partial}{\partial x}\left(\cosh(x)\sin(y)+C(x)\right)=\sinh(x)\sin(y)$$
$$\sinh(x)\sin(y)+C'(x)=\sinh(x)\sin(y)$$
$$C'(x)=0$$
So, in this case $C(x)=C$ ($C$ is any constant, usually it's determined by additional condition, like given function value at some point). The final answer is
$$v=v(x,y)=\cosh(x)\sin(y)+C$$
You can take $C=0$ if you want. Functions $u(x,y)=\sinh(x)\cos(y)$ and $v(x,y)=\cosh(x)\sin(y)+C$ are harmonic conjugates for any $C$.
Differentials $du$ and $dv$ you can calculate using these formulas (and using $(3)$, $(4)$, $(6)$, $(7)$):
$$du = \frac{\partial u}{\partial x}dx + \frac{\partial u}{\partial y}dy$$
$$dv = \frac{\partial v}{\partial x}dx + \frac{\partial v}{\partial y}dy$$
Bonus tip: You won't find function $v$ without using integration operation (at least in general case), if that was the question.
