Integral of a circle to find pi I was thinking of ways to calculate pi, and one idea I had was to integrate $x^2+y^2 = 1$ between 0 and 1 to get $\frac{1}{4}\pi$ Because the total area of such a circle would be $\pi$
I computed this indefinte integral with wolfram alpha:
$\int\sqrt{1-x^2}\ dx$
and got this as a result:
$\frac{1}{2}\Bigr(\sqrt{1-x^2}\ + x + \sin^{-1}(x)\Bigr)+C$
however when I compute the definite integral of
$\int_0^1\sqrt{1-x^2}\ dx$
using
$\frac{1}{2}\Bigr(\sqrt{1-1^2}\ + 1 + \sin^{-1}(1)\Bigr)+C-\Bigr(\frac{1}{2}\Bigr(\sqrt{1-0^2}\ + 0 + \sin^{-1}(0)\Bigr)+C\Bigr)$
45.5 -$\frac{1}{2}$
45
which is obviously not $\frac{1}{4}\pi$
I am aware that 45$^\circ$ converted to radians is $\frac{1}{4}\pi$ but I cannot wrap my head around why the answer to the definite integral is not $\frac{1}{4}\pi$ before an arbitrary conversion of 45 to radians. I would appreciate some help on understanding this.
 A: I believe the fundamental problem here is beyond the fact that you accidentally used degrees instead of radians. There is something deeper here that we need to address. The $\sin$ function that we all know and love from calculus is definitely related to the trigonometric ratio $\sin$, but while they are related, they are not the same thing. Really, they represent different mathematical concepts that just happen to align in a specific context.
In trigonometric studies, the ratio $\sin$ is used in the context of studying right-angled triangles. In this context, the ratio behaves as if it was some function of the angle being considered. But it is not quite a function, because here, angles are explicitly treated as a physical, dimensional quantity, with a specific choice of units. Your choice of units determines what the trigonometric ratio is equal to for any given input. For example, it is clear that $\sin(1^{\circ})$ is not the same as $\sin(1\,\mathrm{rad})$.
So far so good. Where the real confusion lies, though, is in how we talk about the $\sin$ function in pre-calculus and calculus. The $\sin$ function, as a function of real numbers, is not a trigonometric ratio, although how it was historically motivated certainly came from trigonometry. The function is no longer considering angles from a triangle, and so it is not considering inputs with dimension or units: the inputs are now just raw, pure real numbers. In a sense, it even is technically incorrect that we are using radians at all, as our units, because this is not really the case: as an analytical function, it does not even make sense to say that the input has units at all. Yet commonly, we talk about this function as if it just was an extension of the trigonometric ratio, using specifically radians as units. Why is that? Well, first, we need to understand how one defines the $\sin$ function, which is different from how one defines the trigonometric ratio.
Consider the equation of the unit circle in the Cartesian plane. The equation is $x^2+y^2=1$. What we want is a parametrization of this curve, meaning that $x$ and $y$ are functions of a common parameter $t$. We have the initial conditions that $x(0)=1$, and $y(0)=0$, since any other parametrization will just be a shift of the family of parametrizations satisfying these initial conditions. Given a few tedious other restrictions, you can uniquely identify functions $x$ and $y$ that are defined on $\mathbb{R}$, and these functions numerically happen to agree with the trigonometric ratios, given the construction of these functions, but only if those trigonometric ratios are worked in the unit of the radian. This happens because under those heavy restrictoins, the parameter $t$ simply becomes the angle between the ray connecting the origin to the point $(x(t),y(t))$, and the positive $x$-axis, in radians. Hence, these functions act as if they were trigonometric ratios worked on the units of radians, but then they removed the units from the inputs altogether and only kept the raw, dimensionless quantity as real number resulting from that removal. This justifies the naming scheme $x=\cos$ and $y=\sin$, in a common overloading of mathematical notation. However, if one were to consider calculus in a vacuum, devoid of geometry, these functions would still be perfectly well-defined and interesting, and would have the usual properties, and yet one would make no association between them and the trigonometric ratios. This would make the problem of units disappear. In some regard, it is somewhat of a coincidence that radians are the type of unit that make the trigonometric ratios computationally match these functions, even if they are completely different mathematical objects. This is why we speak of the $\sin$ function in the context of calculus as if it was working specifically with radian units. Contrary to what you state in your post, this choice is not arbitrary: it is just the result of a natural alignment between two seemingly-disconnected concepts.
