I'm interested in why $$\int_0^{\pi/2} \sin x\,dx = 1.$$ I know how to do the integral the conventional way but am more interested in what makes radians special for this problem. If we instead compute $$\int_{0}^{90} \sin x^\circ\,dx,$$ we won't get $1$ as the answer.
What about the definition of radians makes this integral evaluate to $1$? I'm looking for an intuitive (presumably geometric) explanation.
Best Answer
I have had a similar question from my friend. He was surprised to find out $\displaystyle \lim_{x^{\circ} \rightarrow 0} \frac{\sin(x^{\circ})}{x^{\circ}} = \frac{\pi}{180}$.
The reason is that we are actually overloading the function name $\sin$. I think the confusion would be subsided if you were to look at $\sin_r(x)$ and $\sin_d(x)$ as two different functions where in the first function you input $x$ in radians and in the second function you input $x$ in degrees and these two are related by $\sin_r(x^r) = \sin_d(y^{\circ})$, where $y^{\circ} =\frac{180}{\pi}x^r$
We know that $\displaystyle \int_{0}^{\pi/2} \sin_r(x^r) dx^r = 1$.
Hence, $\displaystyle \int_{0}^{\pi/2} \sin_d(y^{\circ}) dx^r = 1$, since $\sin_r(x^r) = \sin_d(y^{\circ})$
We have $(\frac{180}{\pi}x^r) = y^{\circ}$. Hence $(\frac{180}{\pi}dx^r) = dy^{\circ}$.
As $x^r$ goes from $0$ to $\pi/2$, $y^{\circ}$ goes from $0$ to $90$.
Hence, we now have $\displaystyle \int_{0}^{90} \sin_d(y^{\circ}) \frac{\pi}{180} dy^{\circ} = 1$.
Hence, $\displaystyle \int_{0}^{90} \sin_d(y^{\circ}) dy^{\circ} = \frac{180}{\pi}$.
EDIT
The question is
"Why is $\int_{0}^{\pi/2} \sin(\theta) d\theta = 1$ when $\theta$ is in radians?"
What follows is my attempt for a purely geometric argument for this.
The geometric argument ultimately hinges (as one would expect) on
"The length of an arc of a circle subtending an angle $\theta$ at the center is $r \theta$ where $\theta$ is in radians"
This essentially comes from the way a radian is defined.
The main crux of the problem is the question
"Why is the average value of $\sin$ over quarter its period is $\frac{2}{\pi}$?"
(Note when we talk of average value of $\sin$ over quarter its period, there is no difference between radians or degrees).
It is easy to see that the average value of $\sin$ over quarter its period lies between $0$ and $1$.
The average value of of $\sin$ over quarter its period is nothing but the average value, with respect to $\theta$, of the height of the line segment as the line segment moves from the right end point of the circle towards the origin such that it is always perpendicular to the $X$ axis, with one end point of the segment on the $X$ axis and the other end point of the segment on the circumference of the circle.
Now here comes the claim. The average value, with respect to $\theta$, of the height of the line segment times the circumference of the quarter of the circle is the area of a unit square.
Let $h$ denote the average value, with respect to $\theta$, of the height of the line segment i.e. $$\displaystyle h = \frac{\int_{0}^{\pi/2} y d \theta}{\int_{0}^{\pi/2} d \theta}$$
The claim is
$$\displaystyle h \times \frac{\pi}{2} = 1$$
Proof:
First note that the area of the square with vertices at $(0,0),(0,1),(1,0),(1,1)$ is obviously $1$.
$\displaystyle \text{Area of square} = 1 = \int_{0}^{1} 1 \times dx = \int_{0}^{1} dx$.
For points on the circle, $\displaystyle x^2 + y^2 = 1 \Rightarrow xdx + ydy = 0 \Rightarrow dy = -\frac{x}{y} dx$.
Further, for the small triangle, we have
$$\displaystyle (dx)^2 + (dy)^2 = (1 \times d \theta)^2$$ $$\displaystyle (dx)^2 + \frac{x^2}{y^2}(dx)^2 = (d \theta)^2$$ $$\displaystyle \frac{x^2 + y^2}{y^2}(dx)^2 = (d \theta)^2$$ $$\displaystyle (dx)^2 = y^2 (d \theta)^2$$
Note that as $\theta$ increases from $0$ to $\frac{\pi}{2}$, $x$ decreases from $1$ to $0$ and hence
$$\displaystyle dx = -y d \theta$$
Hence, $$\displaystyle \text{Area of square} = 1 = \int_{0}^{1} 1 \times dx = \int_{0}^{1} dx = \int_{\pi/2}^0 (-y) d \theta = \int_0^{\pi/2} y d \theta$$
Hence, $\displaystyle 1 = \int_0^{\pi/2} y d \theta = \frac{\int_0^{\pi/2} y d \theta}{\int_0^{\pi/2} d \theta} \times \int_0^{\pi/2} d \theta$
Note that $\displaystyle \int_0^{\pi/2} d \theta = \frac{\pi}{2}$ which is the circumference of the quadrant and $\displaystyle h = \frac{\int_0^{\pi/2} y d \theta}{\int_0^{\pi/2} d \theta}$ is the average value, with respect to $\theta$, of the height of the line segment.
Hence, we get $\displaystyle h \times \frac{\pi}{2} = 1$ i.e. average value, with respect to $\theta$, of the height of the line segment if $\frac{2}{\pi}$.
Hence, we get that $$\displaystyle \int_{0}^{\pi/2} \sin(\theta) d \theta = 1$$ when $\theta$ is in radians