It's not clear to me which is the main question, so I'll try to address most of the questions in a convenient order.
General Background
Throughout, I will suppose that:
- $C$ is parametrized by $a(t)+ib(t)$ (for real $t$ from $0$ to $1$, for convenience) where $x(t)$ and $y(t)$ are real-valued.
- $f(z)=u(z)+iv(z)$ where $u$ and $v$ are real-valued.
Note that
\begin{align*}\int_{C}f(z)\,\mathrm{d}z &= \int_{C}\left(u\left(z\right)+iv\left(z\right)\right)\,\left(\mathrm{d}x+i\mathrm{d}y\right)\\
&=\int_{C}\left(u\left(z\right)\,\mathrm{d}x+iv\left(z\right)\,\mathrm{d}x+iu\left(z\right)\,\mathrm{d}y-v\left(z\right)\,\mathrm{d}y\right)\\
&=\int_{0}^{1}\left(u\left(a(t)+ib(t)\right)a'(t)+iv\left(a(t)+ib(t)\right)a'(t)\right.\\
&\phantom{=}\left.+iu\left(a(t)+ib(t)\right)b'(t)-v\left(a(t)+ib(t)\right)b'(t)\right)\,\mathrm{d}t\end{align*}
Conjugating the curve
Now, if we replace the curve $C$ by the conjugate curve $\overline{C}$ parametrized by $a(t)-ib(t)$, then we have:
\begin{align*}\int_{\overline{C}}f(z)\,\mathrm{d}z &= \int_{\overline{C}}\left(u\left(z\right)+iv\left(z\right)\right)\,\left(\mathrm{d}x+i\mathrm{d}y\right)\\
&=\int_{\overline{C}}\left(u\left(z\right)\,\mathrm{d}x+iv\left(z\right)\,\mathrm{d}x+iu\left(z\right)\,\mathrm{d}y-v\left(z\right)\,\mathrm{d}y\right)\\
&=\int_{0}^{1}\left(u\left(a(t)\boxed{-}ib(t)\right)a'(t)+iv\left(a(t)\boxed{-}ib(t)\right)a'(t)\right.\\
&\phantom{=}\left.\boxed{-}iu\left(a(t)\boxed{-}ib(t)\right)b'(t)\boxed{+}v\left(a(t)\boxed{-}ib(t)\right)b'(t)\right)\,\mathrm{d}t\end{align*}
There are a few ways to get to this output by modifying $\int_{C}f(z)\,\mathrm{d}z$. Certainly, changing the argument of $f$ from $z$ to $\overline{z}$ would be part of a fix. Aside from that, we can conjugate the output of $f$ to flip the signs of the two $v$ terms (which makes the $v$ term match but also makes the $iv$ term wrong), and then conjugate the output of the integral to flip the signs of the two $i$ terms. In symbols, we have:
\begin{align*}\overline{\int_{C}\overline{f(\overline{z})}\,\mathrm{d}z}&=\overline{\int_{C}\left(u\left(\overline{z}\right)-iv\left(\overline{z}\right)\right)\,\left(\mathrm{d}x+i\mathrm{d}y\right)}\\&=\overline{\int_{C}\left(u\left(\overline{z}\right)\,\mathrm{d}x-iv\left(\overline{z}\right)\,\mathrm{d}x+iu\left(\overline{z}\right)\,\mathrm{d}y+v\left(\overline{z}\right)\,\mathrm{d}y\right)}\\&=\int_{C}\left(u\left(\overline{z}\right)\,\mathrm{d}x+iv\left(\overline{z}\right)\,\mathrm{d}x-iu\left(\overline{z}\right)\,\mathrm{d}y+v\left(\overline{z}\right)\,\mathrm{d}y\right)\\&=\int_{0}^{1}\left(u\left(a(t)-ib(t)\right)a'(t)+iv\left(a(t)-ib(t)\right)a'(t)\right.\\&\phantom{=}\left.-iu\left(a(t)-ib(t)\right)b'(t)+v\left(a(t)-ib(t)\right)b'(t)\right)\,\mathrm{d}t\\&=\int_{\overline{C}}f(z)\,\mathrm{d}z\checkmark\end{align*}
To emphasize:
\begin{equation}\int_{\overline{C}}f(z)\,\mathrm{d}z=\overline{\int_{C}\overline{f(\overline{z})}\,\mathrm{d}z}\tag{1}\end{equation}
(Note that this is very similar to copper.hat's answer.)
Integration with Respect to $\overline{z}$
General Definitions
What is $\int\limits_{C}\left(f(z,\bar{z})\,dz+g(z,\bar{z})\,d\bar{z}\right)$?
Just as $\mathrm dz$ is essentially a shorthand for $\mathrm dx+i\mathrm dy$, similarly $\mathrm d\overline z$ is a shorthand for $\mathrm dx-i\mathrm dy=\overline{\mathrm dz}$.
This allows us to write, for instance:
\begin{align}
&\phantom{=} \int_{C}f(z,\overline{z})\,\mathrm{d}z+g(z,\overline{z})\,\mathrm{d}\overline{z}\\
&=\int_{C}f(z,\overline{z})\,\left(\mathrm{d}x+i\mathrm{d}y\right)+g(z,\overline{z})\,\left(\mathrm{d}x-i\mathrm{d}y\right)\\
&=\int_{C}f(x+iy,x-iy)\,\left(\mathrm{d}x+i\mathrm{d}y\right)+g(x+iy,x-iy)\,\left(\mathrm{d}x-i\mathrm{d}y\right)\\
&=\int_{0}^{1}\left(f\left(a(t)+ib(t),a(t)-ib(t)\right)\cdot\left(a'(t)+ib'(t)\right)\right.\\
&\phantom{=}\left.+g\left(a(t)+ib(t),a(t)-ib(t)\right)\cdot\left(a'(t)-ib'(t)\right)\right)\,\mathrm{d}t
\end{align}
If you'd like, you may separate the outputs of $f$ and $g$ into their real and imaginary parts, so that you could distribute and simplify the above to find real integral expressions for its real and imaginary parts.
more general questions about complex 1-forms?
For some more about those outside of the context of integration, see, for instance, my answer to Why could we use the complex basis for the cotangent space and what does Cauchy Integral Formula in this basis say?.
Conversion to $\int\cdot\,\mathrm dz$
how exactly do we formulate $\int\limits_{C}f(z,\bar{z})\,d\bar{z}$ as an integral integrated $dz$?
Ignoring the $\bar{z}$ argument since it would complicate things and could be addressed in a separate question, $\mathrm{d}\overline{z}=\mathrm{d}x-i\mathrm{d}y$ yields:
\begin{align*}\int_{C}f(z)\mathrm{d}\overline{z}&=\int_{C}\left(u\left(z\right)+iv\left(z\right)\right)\left(\mathrm{d}x\boxed{-}i\mathrm{d}y\right)\\&=\int_{C}\left(u\left(z\right)\mathrm{d}x+iv\left(z\right)\mathrm{d}x\boxed{-}iu\left(z\right)\mathrm{d}y\boxed{+}v\left(z\right)\mathrm{d}y\right)\\&=\int_{0}^{1}\left(u\left(a(t)+ib(t)\right)a'(t)+iv\left(a(t)+ib(t)\right)a'(t)\right.\\&\phantom{=}\left.\boxed{-}iu\left(a(t)+ib(t)\right)b'(t)\boxed{+}v\left(a(t)+ib(t)\right)b'(t)\right)\mathrm{d}t\end{align*}
To obtain another formula, we can use a similar double-conjugation trick to what worked to obtain equation (1):
\begin{align*}\overline{\int_{C}\overline{f(z)}\,\mathrm{d}z}&=\overline{\int_{C}\left(u\left(z\right)\boxed{-i}v\left(z\right)\right)\,\left(\mathrm{d}x+i\mathrm{d}y\right)}\\&=\overline{\int_{C}\left(u\left(z\right)\,\mathrm{d}x\boxed{-}iv\left(z\right)\,\mathrm{d}x+iu\left(z\right)\,\mathrm{d}y\boxed{+}v\left(z\right)\,\mathrm{d}y\right)}\\&=\int_{C}\left(u\left(z\right)\,\mathrm{d}x\boxed{+}iv\left(z\right)\,\mathrm{d}x\boxed{-}iu\left(z\right)\,\mathrm{d}y+v\left(z\right)\,\mathrm{d}y\right)\\&=\int_{0}^{1}\left(u\left(a(t)+ib(t)\right)a'(t)+iv\left(a(t)+ib(t)\right)a'(t)\right.\\&\phantom{=}\left.-iu\left(a(t)+ib(t)\right)b'(t)+v\left(a(t)+ib(t)\right)b'(t)\right)\mathrm{d}t\\&=\int_{C}f(z)\mathrm{d}\overline{z}\checkmark\end{align*}
To emphasize:
\begin{equation}\int_{C}f(z)\mathrm{d}\overline{z}=\overline{\int_{C}\overline{f(z)}\,\mathrm{d}z}\tag{2}\end{equation}
Other identities
A Comprehensive Table
By applying equations (1) and/or (2) with the functions $f$, $z\mapsto f(\overline{z})$, $z\mapsto \overline{f(z)}$, and $z\mapsto \overline{f(\overline{z})}$, we can obtain a whole host of identities:
\begin{array}{rc|rc}
\int_{C}f(z)\,\mathrm{d}z=: & a & \int_{C}f(z)\,\mathrm{d}\overline{z}= & \overline{c}\\
\int_{C}f(\overline{z})\,\mathrm{d}z=: & b & \int_{C}f(\overline{z})\,\mathrm{d}\overline{z}= & \overline{d}\\
\int_{C}\overline{f(z)}\,\mathrm{d}z=: & c & \int_{C}\overline{f(z)}\,\mathrm{d}\overline{z}= & \overline{a}\\
\int_{C}\overline{f(\overline{z})}\,\mathrm{d}z=: & d & \int_{C}\overline{f(\overline{z})}\,\mathrm{d}\overline{z}= & \overline{b}\\
\hline \int_{\overline{C}}f(z)\,\mathrm{d}z= & \overline{d} & \int_{\overline{C}}f(z)\,\mathrm{d}\overline{z}= & b\\
\int_{\overline{C}}f(\overline{z})\,\mathrm{d}z= & \overline{c} & \int_{\overline{C}}f(\overline{z})\,\mathrm{d}\overline{z}= & a\\
\int_{\overline{C}}\overline{f(z)}\,\mathrm{d}z= & \overline{b} & \int_{\overline{C}}\overline{f(z)}\,\mathrm{d}\overline{z}= & d\\
\int_{\overline{C}}\overline{f(\overline{z})}\,\mathrm{d}z= & \overline{a} & \int_{\overline{C}}\overline{f(\overline{z})}\,\mathrm{d}\overline{z}= & c
\end{array}
Note that this table includes cases such as $\int_{C}\overline{f(z)}\mathrm{d}\overline{z}=\overline{\int_{C}f(z)\,\mathrm{d}z}$ from Sangchul Lee's answer and
Why should $\int\limits_{\gamma}f(z)\,dz=\int\limits_{\bar{\gamma}}f(\bar{z})\,d\bar{z}$?
the comment you mentioned.
Replacing $i$ with $-i$
replacing all instances...this integral must equal $−i$...Or am I making some kind of mistake?
Now that we have the above table of identities, we can easily see what options we have to get to $\overline{\int_{C}f(z)\,\mathrm{d}z}=\overline{a}$. Specifically, we could use one of:
- $\int_{C}\overline{f(z)}\,\mathrm{d}\overline{z}$
- $\int_{\overline{C}}\overline{f(\overline{z})}\,\mathrm{d}z$
- $\overline{\int_{\overline{C}}f(\overline{z})\,\mathrm{d}\overline{z}}$.
The third expression is not satisfying because it has conjugation on the outside.
The first expression is odd because it may seem like a coincidence that an "anti-contour integral" (integration with respect to $\overline{z}$) of the conjugate function equals the conjugate of the original integral; it doesn't seem to capture the intent of "[replacing] each $i$ with a $−i$".
But I believe the second expression is just right: We keep the $\mathrm dz$ since we still want to do a contour integral, but we have conjugated the curve and the input and the output of the function. Since we've basically conjugated everything in sight, it makes sense that the result is the conjugate of the original integral.
(Note that in the specific and subtle case of $f(z)=z$ that appeared in the OP, we have $\int_{\overline{C}}\overline{f(\overline{z})}\,\mathrm{d}z=\int_{\overline{C}}\overline{\overline{z}}\,\mathrm{d}z=\int_{\overline{C}}z\,\mathrm{d}z$.)
