Lets say I have
$\int{ \overrightarrow{B} \cdot\ \overrightarrow{dA} }$
is that equal to
$ \int{ B \cdot dA } \cdot\cos(\theta)$
or
$\int{(B \cdot\cos(\theta)) \cdot dA} $
For example: a magnetic flux on a rotating surface, where the magnetic field is uniform and the area vector is rotating.
In the case where the $\cos(\theta)$ is inside the integral, theta is changing, so the integral wants to see the theta put into terms of A or the other way around; Whereas in the case where $\cos(\theta)$ is outside of the integral, B is uniform (constant with respect to space) can be pulled out, and the integral(dA) can be evaluated as A therefore $\Phi_B = BA\cos(\theta)$.
Edit: In the example $\theta$ is known- such as a rotating solenoid where the angle between the magnetic vector and the normal vector of the surface area lie in the same plane. Ex: $\theta = \omega \cdot t$ and both $\omega$ and t are known.
This type of integral is known as a line integral. Here is the process by which you would go about this type of problem.
Suppose we are given some vector field $F:\mathbb{R}^n\to\mathbb{R}^n$ and some path $\gamma$ in $\mathbb{R}^n$. The line integral will tell us how much the path lines up with the vector field. In physics, this is often used for summing up a force field along the path, yielding the energy gained by traveling along that path. Now, we look at the line integral:
$$\int_\gamma F\cdot \mathrm{d}r$$
Note that $\mathrm{d}r=\begin{pmatrix}\mathrm{d}x^1\\\mathrm{d}x^2\\\vdots\\\mathrm{d}x^n\end{pmatrix}$.
This is summing up the projections of $F$ onto $\gamma$ across all of $\gamma$. Now, we parameterize $r$ as $r:[0,1]\to\mathbb{R}^n$ such that $r([0,1])=\gamma$, and denote this parameterization as $r(t)$. Note that this means all the components of $r(t)$ are functions of $t$. Now, by the multivariable chain rule:
$$\mathrm{d}r=r'(t)\hspace{1mm}\mathrm{d}t$$
Thus:
$$\int_\gamma F\cdot \mathrm{d}r=\int_0^1F(r(t))\cdot r'(t)\hspace{1mm}\mathrm{d}t$$
Which can then be solved using standard single variable calc techniques.
EDIT
The process described above is for a line integral, however, in the OP's question, the integral given was a surface integral.
Suppose we are given some force field $F:\mathbb{R}^n\to\mathbb{R}^3$ and either a $2$-cell or $3$-cell $A$ (a subset of $\mathbb{R}^3$ whose topological interior is diffeomorphic to $\mathbb{B}^2$ or $\mathbb{B}^3$ resp.). Then, the surface integral of $F$ over $A$ is defined to be:
$$\int_AF\cdot\hat{n}\hspace{1mm}\mathrm{d}s$$
Let's break this down, both intuitively and rigorously.
INTUITIVELY
We are summing up the magnitude of the parts of $F$ normal to $A$. The $F\cdot\hat{n}\hspace{1mm}\mathrm{d}s$ part of the integral is the inner product of $F$ with the unit vector normal to $\mathrm{d}s$, where $\mathrm{d}s$ is a small, "flat" piece of $A$. The $\int_A$ means we are summing up these magnitudes across the entire $2$-cell $A$.
RIGOROUSLY
We then define a smooth parameterization of $A$: $a(u,v):I\times I\to\mathbb{R}^3$, which has the property $a(I\times I)=A$. Now, we can define what we mean by $\hat{n}\hspace{1mm}\mathrm{d}s$ by using the cross product and a type of mapping called a differential $2$-form (basically a way of representing area):
$$\hat{n}\hspace{1mm}\mathrm{d}s=\frac{\frac{\partial a}{\partial u}\times\frac{\partial a}{\partial v}}{||\frac{\partial a}{\partial u}\times\frac{\partial a}{\partial v}||}\hspace{1mm}\mathrm{d}u\wedge\mathrm{d}v=\frac{\frac{\partial a}{\partial u}\times\frac{\partial a}{\partial v}}{||\frac{\partial a}{\partial u}\times\frac{\partial a}{\partial v}||}\hspace{1mm}\mathrm{d}u\hspace{1mm}\mathrm{d}v$$
Thus, our surface integral may be written:
$$\int_AF\cdot\hat{n}\hspace{1mm}\mathrm{d}s=\int_0^1\int_0^1F\cdot\frac{\frac{\partial a}{\partial u}\times\frac{\partial a}{\partial v}}{||\frac{\partial a}{\partial u}\times\frac{\partial a}{\partial v}||}\hspace{1mm}\mathrm{d}u\hspace{1mm}\mathrm{d}v$$
There are then a couple of theorems that can help to solve this:
GAUSS'S THEOREM
Suppose that $A$ is a $3$-cell. Then, $\partial A$ is diffeomorphic to the sphere $\mathbb{S}^2$ (this is to help with intuition), and:
$$\int_{\partial A}F\cdot\hat{n}\hspace{1mm}\mathrm{d}s=\int_A\text{div}\hspace{.5mm}F\hspace{1mm}\mathrm{d}V$$
STOKES'S THEOREM
Suppose that $A$ is a $2$-cell. Then:
$$\int_A\text{curl}\hspace{.5mm}F\cdot\hat{n}\hspace{1mm}\mathrm{d}s=\int_{\partial A}F\cdot\mathrm{d}r$$
Where we can solve this line integral in the way explained above.
CONCLUSION
The reason we do not use $F\cdot\hat{n}\hspace{1mm}\mathrm{d}s=(\cos\theta)\hspace{.5mm}|F|\hspace{.5mm}|\hat{n}\hspace{1mm}\mathrm{d}s|$ is because we do not know the angle between the tangent space at that point and the field $F$. Additionally, there is not much meaning to $|\hat{n}\hspace{1mm}\mathrm{d}s|=|\hat{n}|\hspace{.5mm}|\mathrm{d}s|=|\mathrm{d}s|$, as $\mathrm{d}s$ is a differential representing a tiny bit of area.