Is there a useful notion of complex area? I am coming to the end of my first year at university, in case that's helpful.
There are two ways in which I have heard of area being defined, a real-valued scalar which represents the 2 dimensional size of an object and a vector which lies perpendicular to a surface whose magnitude defines the size. Both of these are, as far as I'm aware, an Integral of some function that defines the boundaries of the object. I have a good knowledge of what complex numbers are and have used complex-valued functions a fair amount. If we take a relatively simple integral like
$$\int \cos(x) + \dot{\imath}  \sin(x) dx = \sin(x) - \dot{\imath} \cos(x) $$
Then we get a complex answer. This is usually true because of: $\int f(x) + g(x) = \int f(x) + \int g(x)$ and $\int a f(x) = a \int f(x)$ where in this case i is a.
In this situation it seems clear to me that the area between the function and the input axis (assuming its a real-valued input range) is no longer represented by this integral. Even before you reach its periodicity at $2\pi$ it seems to me that area is just equal to $x$.
$$\int \sqrt{sin^2(x)+cos^2(x)} dx =\int \sqrt{1} dx = x$$
And I can't think at all where or how you would represent the area of this spiral as a vector.
Now, there is the notion of "anti-area" where areas below the x-axis are counted as negative, and I assume this holds true for complex numbers in that areas in the $+\dot{\imath}$ direction are 
counted as imaginary and areas in the $-\dot{\imath}$ direction are 
negative imaginary. So say we have a function that stays strictly non-negative in both its real and imaginary parts. My first thought was just to have the unit vector wiggle about in the positive quadrant like this:

Using the equation $e^{i sin(x)/2 + \pi/4 }$. In a case like this, where there is no canceling out and we get a value of $4.16951 + 4.16951i$ for the definite integral over the course of a period.


*

*Is there an obvious geometric interpretation to the value $4.16951 + 4.16951i$ that I'm just too dumb to see yet?

*Are there functions too irreducably complex for a non-complex area to even make sense? i.e. Are there situations in which representing areas with complex numbers is neccesary?

*Should I just give up on the notion of integrals as area and be content with them representing the inverse of a derivative?
 A: I'll answer your last question first, since I think it informs what I'll say next. While their interpretation as areas and relationship to derivatives is extremely important, a priori integrals have nothing to do with areas or antiderivatives; they're defined (for well-behaved functions) as Riemann sums. For (positive) real functions $f:\mathbb{R}\to\mathbb{R}$, this has the immediate interpretation as "approximating the area by rectangles", as shown in this graphic from Wikipedia.

In other situations, there might not be such a natural interpretation of $\int f\, dx$ as an area. In general, you should think of the integral as a Riemann sum which essentially involves dividing your domain $D$ up into $N$ small "elements" $\Delta_i$ (in this case small intervals), assigning a value $F_{\Delta_i}$ to each $\Delta_i$ using the function (for example, the value at one of the endpoints or in the middle, the maximum or minimum value...), taking the sum $\sum_{i=n}^N F_{\Delta_i}\Delta_i$ and taking some kind of limit where the $\Delta_i$s become arbitrarily small and the number of them goes to infinity. If this limit is finite and agrees for any sensible way of assigning the values $F_{\Delta_i}$, you call it $\int f_D\,dx$.
Now, taking your first example, we have a function $f:\mathbb{R}\to\mathbb{C}$ with $f(x)=\cos(x)+i\sin(x)=e^{ix}$; I can see 2 different ways of interpreting what this function means and these give different interpretations of the integrals you showed.
For simplicity I assume that all of the functions I mention are smooth and differentiable.
Interpretation 1: Forget about the complex numbers
Since $i$ doesn't really come into play here except as a unit vector, we can think of $\mathbb{C}$ as $\mathbb{R}^2$ (via thinking about the Argand diagram) and treat $f$ as a function $\vec{f}:\mathbb{R}\to\mathbb{R}^2$ defined by $\vec{f}(x)=(\cos(x),\sin(x))$. I think it's then natural to plot this function in 3 dimensions as you did with your other function. What about the integrals then? The first integral you wrote would be $(\int\cos(x)\,dx,\int\sin(x)\,dx)=(\sin(x),-\cos(x))$; I'm not sure that there's a geometric/ physical interpretation to this, to be honest, and I don't know where you would encounter such an object. 
On the other hand, $|\vec{f}(x)|=\sqrt{\cos^2(x)+\sin^2(x)}=1$ gives the distance of $\vec{f}(x)$ from $0$. Then $\int|\vec{f}(x)|\,dx$ is the Riemann sum of such distances; the interpretation is that
  $$ \frac{1}{(b-a)} \int^b_a|\vec{f}(x)|\,dx$$
is the mean distance of $\vec{f}(x)$ from $0$ when $x\in[a,b]$. In this case this is just $1$, of course. Another computation you could do is
  $$ \int^b_a  \left|\frac{d}{dx}\vec{f}(x)\right|\,dx 
    = \int^b_a \left|(-\sin(x),\cos(x))\right|\,dx
    = \int^b_a 1\,dx
    = (b-a)$$
which gives the arc length of the path traced out $\vec{f}(x)$ as you vary $x$. If we consider $\frac{d}{dx}\vec{f}(x)$ to be a velocity ($x$ now has the role of time), $|\frac{d}{dx}\vec{f}(x)|$ is a speed and so this integral is just the "distance travelled" along the graph of $\vec{f}(x)$.
For more about integrals along (planar) curves and their relationship to area, I recommend you read about line integrals and Green's Theorem.
Interpretation 2: Embrace the complexity
This is actually quite similar to what happened above, but in the complex plane it generalises in a beautiful way which makes complex functions seem magical. I think that a 3d plot in this case isn't greatly helpful for understanding what's going on. Instead, we do the following.
We define a path as a function $\gamma:\mathbb{R} \to \mathbb{C}$. We think about $\mathbb{R}$ as a timeline and $\gamma(t)$ as a point in $\mathbb{C}$ at the time $t$. Now we have a film where over an interval (a,b), $\gamma$ traces out a curve or contour $C\subset \mathbb{C}$. We can use integrals as we did in $\mathbb{R}^2$ to talk about average distance from the origin and arc length of $\gamma$, but something more interesting can happen here.
The notion of complex integral which is usually used is that of the contour integral. For a complex function $f:\mathbb{C}\to\mathbb{C}$ over the contour $C$, this is denoted
    $$ \int_C f(z)\,dz. $$
This can be defined using Riemann sums again, breaking up $C$ into small segments, assigning values to each segment using $f$ etc. This is equivalent to the following more practical definition. We parametrise $C$ using $\gamma$ - there might be paths other than $\gamma$ which trace out $C$ but it turns out that it doesn't matter which we choose. Then
  $$ \int_C f(z)\,dz 
= \int^b_a f(\gamma(t))\frac{dz}{dt}\,dt
= \int^b_a \Re\left(f(\gamma(t))\frac{dz}{dt}\right)\,dt 
  + i\int^b_a \Im\left(f(\gamma(t))\frac{dz}{dt}\right)\,dt,$$
where the two integrals on the RHS are real so we know how to solve them. Now, this doesn't have any interpretation as an area, but something interesting can happen. 
In your first example, we can take $\gamma$ to be the path $\gamma(t)=t$, where we think of $t=t+0i$ on the RHS. Here the curve $C$ will just be a segment of the real line $\mathbb{R}\subset \mathbb{C}$. We take $f(z):=e^{iz}$. Then we get the integral you found, with appropriate limits of integration. We could instead have done $\gamma(t)=e^{it}$ and $f(z)=z$ to get the same answer, but we have computed something slightly different which leads to something interesting. Now, if we restrict to $t\in(0,\theta)$ where $0<\theta\leq 2\pi$, $C$ is a segment of the unit circle in the complex plane and we're integrating the function $f(z)$. If we take $\theta=2\pi$, the contour closes and the definite integral is $0$. 
An amazing fact this that if you take any complex differentiable function $f$ and any closed contour $C$ which doesn't cross over itself, $\int_C f(z)\,dz=0$. Even better, if $f$ is differentiable apart from at a discrete set of singularities, the value of $\int_C f(z)\,dz$ only depends on data having to do with those singularities, their so-called residues. This is Cauchy's Residue Theorem. For example, 
 $$\int_C \frac{1}{z}\,dz = \begin{cases}2\pi i & \text{if $C$ encloses $0$},\\ 0 & \text{otherwise}.\end{cases}$$
One interpretation of this is that contour integrals of complex differentiable functions are really just telling you about their singularities. This is highly non-intuitive and non-obvious, but it is true and rather beautiful. It's also extremely useful and is applied in all sorts of situations, from solving difficult real integrals to predicting the properties and behaviour of particles in Quantum Field Theory.
Final comments
I'll quickly address some the other points you brought up. In general it doesn't make sense to think about area as a vector. When calculating the area of a curved surface, or an integral along a surface (such as the flux of a magnetic field through a surface in physics), the "elements" which go into the Riemann sum can be be thought of as vectors normal to a little flat square which approximates the surface in a small region, with magnitude given by the area of that little square. When we do the integral however, we just get a scalar.
The number you get when you integrate $e^{i\sin(x)+\frac{\pi}{2}}$ has no geometric interpretation which is apparent to me, and it probably doesn't have much meaning for the following reason: it could be re-interpreted as a contour integral of $f(z)=e^z$ on a contour given by $\gamma(t)=i\sin(t)+\frac{\pi}{2}$. Cauchy's Residue Theorem tell us that the contour we take doesn't actually matter; the integral only depends on the endpoints $\gamma(a)$, $\gamma(b)$.
