why radians can be converted to reals in calculus? Consider this integral:
$$ \int \sin^2x dx = \frac x2 - \frac {\sin2x}4 + C $$
Note the first term $\frac x2$ is a real as opposed to radian and can, in fact, be substituted with a real number when taking definite integral.
To make the statement more clear, introduce trigonometric derivatives in degree form:
$$ \frac {d}{dx} \sin^\circ x = \frac \pi {180} \cos^\circ x  $$
However, this does not change the frist term of the integral...
$$ \int \sin^{\circ2}xdx = \int \frac 12 - \frac {\cos^\circ2x}2
= \frac x2 - \frac {180}\pi \times \frac {\sin^\circ2x}4 + C $$
Then in this content, what is $\frac x2$, real or radian?
 A: The variable $x$ has no units. It depends on how you use this integral, and what you're modelling. It's possible, for example, that you have some use for putting the angle, in radians, into the function $\sin^\circ$, for totally normal reasons. It's not likely, but it can happen.
However, such situations are overwhelmingly unlikely, and the use of $\sin^\circ$ is overwhelmingly suggestive that, in every physical use of this particular integral, you will get erroneous results from substituting values in for $x$ in any angle measure other than degrees.
A: Your distinction between "reals" and "radians" is not a meaningful one.  Radians are a unitless measurement, so "$x$ radians" is, in fact, just the real number $x$ (understood in a particular context---that of angles).
Now, $\sin(x)$ and $\sin^\circ(x)$ are both functions from $\mathbb{R}$ to $\mathbb{R}$, but they are DIFFERENT functions!  Notice that $\sin(x)$ has period $2\pi$, but $\sin^\circ(x)$ has period 360.  Yes, a term of $x/2$ appears in the antiderivative of both and there is a different "scaling" factor in front of the second term (but again, the second term is different between the two examples because $\sin(x)$ is not the same function as $\sin^\circ(x)$).  But this doesn't mean anything a priori---the functions are different, so their integrals will be different.  Of course, the form of the antiderivative is similar because the two functions are related by $\sin^\circ(x) = \sin(180x/\pi)$.  Thats all that's really going on.
