Finding $\int_0^{\pi/2} \sin x\,dx$

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

Think of the definite integral as giving displacement, and use the fact that the key feature of radians as an angle measure is that the arc length of the unit circle subtended by an angle of radian measure $x$ is $x$ units.

More specifically, imagine you are traveling along the unit circle at a speed of 1 rad/s starting at (-1,0) and moving clockwise. Since you are in radians, this means that you are traveling 1 m/s (to pick a distance unit) on the circle. When you have traveled $x$ units, your position vector is $(-\cos x, \sin x)$. The velocity vector must be tangent to the circle and thus is perpendicular to the position vector. Thus the velocity vector is in the direction of $(\sin x, \cos x)$. Because the speed is 1 m/s, the velocity vector actually is $(\sin x, \cos x)$ and not some other scalar multiple of it. Thus $\int_0^{\pi/2} \sin x dx$ when $x$ is in radians is just the horizontal displacement on the unit circle when traveling $\pi/2$ radians along the circle's perimeter from (-1,0) to (0,1). Since the horizontal displacement is 1, $\int_0^{\pi/2} \sin x dx = 1$.

Note the effect of radians in making the speed 1 m/s. If you were traveling at 1 degree/s, the speed along the circle would be $\frac{\pi}{180}$ m/s, and you would have $\int_0^{90} \sin x \frac{\pi}{180} dx = 1$ as in Sivaram's answer.