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?

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.
