Here's how to find the corresponding imaginary parts by educated guessing. Maybe not the most systematic method, but maybe it improves your intuition about how these things behave.
For the first, remember that $e^{a + ib} = e^a(\cos b + i\sin b)$ (assuming $a,b \in \mathbb{R}$). In other words, $e^z$ sort of maps from polar coordinates to cartesian coordinates. $\log(z)$ thus does the reverse mapping, i.e. $\log(a+ib) = \log(|z|) + i\arg(a+ib) + i2\pi n$. This motivates the assumption that your first $u$ is the real part of $$
f(z) = \log(z^2)
$$
Which it actually is, since $\Re(f(x+iy)) = \log(|(x+iy)^2|) = \log(|x+iy|^2) = log(x^2+y^2)$. The corresponding $v$ is thus $v(x,y) = \Im(f(x+iy))$, i.e. $$
v(x,y) = \arg(z^2) + 2\pi n= 2\arg(z) + 2\pi n= 2\left(arctan\left(\frac{y}{x}\right) + \pi \tilde{n}\right)
$$
Your integration yields the wrong result because you forgot to take the denominator (which depends on $y$!) into account when finding the antiderivative.
For the second, use that $\sin(x+iy)$ = $\sin(x)\cos(iy) + \cos(x)\sin(iy)$ = $\sin(x)\cosh(y) +i\cos(x)\sinh(y)$ . From that, you get that $$
f(z) = \sin(z)
$$
and $$
v(x,y) = \cos(x)\sinh(y)
$$
which is the same as your integration yields.
For the third, observe that $x^2+y^2 = z\bar{z}$ if $z = x+iy$. Thus, one guess for $f(z)$ could be $\frac{z}{z\bar{z}}$. That's no good, however, because $z \to \bar{z}$ is not holomorphic. But since the real parts of $z$ and $\bar{z}$ are identical, you can also try $$
f(z) = \frac{\bar{z}}{z\bar{z}} = \frac{1}{z}
$$ And voilá, since $$
\frac{1}{z} = \frac{1}{x+iy} = \frac{x-iy}{(x+iy)(i-iy)} = \frac{x - iy}{x^2 + y^2}
$$
you indeed have $\Re(f(x+iy)) = \frac{x}{x^2+y^2}$, and thus $$
v(x,y) = \frac{-y}{x^2+y^2}
$$
So am I going to try my best to answer this, don't be too harsh on me if it is all wrong!
I think you will want to look at the Cauchy-Riemann equations in polar coordinates, namely:
$$\frac{\partial u}{\partial r} = \frac{1}{r} \frac{\partial v}{\partial \theta}$$ and $$\frac{ \partial v}{\partial r} = - \frac{1}{r} \frac{\partial u}{\partial \theta}$$
The reason I say this is because we are going to be integrating using polar coordinates. What we are going to do is given our harmonic function $u(r,\theta)$ we will take partial derivatives with respect to $r$ and $\theta$ giving us $u_r$ and $u_\theta$ respectively. We will then solve for $v$ using the Cauchy Riemann equations and integrating.
To this end, set
$$v(r,\theta) = \int_a^r -\frac{1}{t} u_\theta(t,0) dt + \int_0^{\theta} r u_r(r,\psi) d\psi$$
We can then check that $u$ and $v$ satisfy the Cauchy-Riemann equations by taking partial derivatives.
We get:
$$\frac{\partial v}{\partial \theta} = r u_r = r \frac{\partial u}{\partial r}$$
by the Fundamental Theorem of calculus and
$$\frac{\partial v}{\partial r} = - \frac{1}{r}u_\theta(r, 0) + \int_0^\theta \frac{\partial}{\partial r} r u_r (r, \psi) d\psi$$
(we can differentiate under the integral sign because $u$ is $C^2$ and the region is compact)
$$= \int_0^\theta u_r(r,\psi)d\psi + \int_0^\theta r \frac{\partial^2 u}{\partial r^2} (r, \psi) d\psi$$
Now, $u$ is harmonic, so the Laplace equation in polar coordinates gives us
$$\frac{\partial^2 u}{\partial r^2} = -\frac{1}{r}\frac{\partial u}{\partial r} -\frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2}$$
Hence, the above is
$$= \int_0^\theta u_r(r,\psi)d\psi + \int_0^\theta r \left( -\frac{1}{r}u_r(r,\psi) -\frac{1}{r^2}u_{\theta\theta}(r,\psi) \right ) d\psi$$
$$=\int_0^\theta u_r(r,\psi)d\psi - \int_0^\theta u_r(r,\psi)d\psi + \int_0^\theta -\frac{1}{r}u_{\theta\theta}(r,\psi) d\psi= -\frac{1}{r}u_\theta$$
Hence, $$\frac{\partial v}{\partial r} = -\frac{1}{r}\frac{\partial u}{\partial \theta}$$
So $u,v$ satisfy the Cauchy Riemann equations.
Best Answer
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.