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.

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    $\begingroup$ If you measure in degrees instead of radians, the derivative of $\cos(x)$ is not $-\sin(x)$; you're off by a constant (think the graph: instead of having a period of $2\pi$, it has a period of $360$, almost 60 times longer). So $\cos(x)$ is no longer an antiderivative of $\sin(x)$. If you do the integral with the correct antiderivative, you wll get $1$. $\endgroup$ – Arturo Magidin Feb 13 '11 at 5:52
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    $\begingroup$ Mathematica says (and I'm inclined to agree) that $\int_0^{\pi/2} \sin x\,dx =\int_{0^\circ}^{90^\circ} \sin x\,dx= 1$, whereas $\int_{0}^{90} \sin x^\circ\,dx=\frac{1}{1^\circ}=\frac{180}{\pi}$ (though this is more of an issue of notation than actual substance). $\endgroup$ – Isaac Feb 13 '11 at 6:05
  • $\begingroup$ @Isaac, you're right that that does make more sense. I'll change my question (though I'm not sure it was particularly unclear originally). $\endgroup$ – Sophie Alpert Feb 13 '11 at 19:58
  • $\begingroup$ @Arturo: Yes, but why is the derivative of $\cos(x)$ equal to $-\sin(x)$ when $x$ is measured in radians? $\endgroup$ – Sophie Alpert Feb 13 '11 at 19:59
  • $\begingroup$ @Ben: Because $$\lim_{x\to 0}\frac{\sin x}{x} = 1$$when $x$ is measured in radians. Basically, because radians give the arclength parametrization of the unit circle (one radian yields one unit of arc length), and degrees do not. Radians are the "normalized" way of measuring angles, so that pesky proportionality constants all become $1$ when you use radians. $\endgroup$ – Arturo Magidin Feb 13 '11 at 20:02

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, $\displaystyle \int_{0}^{90} \sin_d(y^{\circ}) dy^{\circ} = \frac{180}{\pi}$.


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.

enter image description here

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. enter image description here

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

$$\displaystyle h \times \frac{\pi}{2} = 1$$


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

  • $\begingroup$ @Ben: Good Question. To think of it in a possibly geometric setting, we will make a very crude approximation. The average value of the sin function over half a period is $\approx \frac{1}{2}$ (half of the extreme values $0$ and $1$). Now if you evaluate the integral in radians from $0$ t0 $\frac{\pi}{2}$, you will get $\frac{1}{2} \frac{\pi}{2} = \frac{\pi}{4} \approx 0.785$. Now if you were to evaluate the integral in degrees from $0$ to $90$, you will get $\frac{1}{2} 90 = 45$. The exact values of the integrals are $1$ and $\frac{180}{\pi} \approx 57$ respectively. $\endgroup$ – user17762 Feb 13 '11 at 6:09
  • $\begingroup$ So you answered why the integrals give different answers scaled by a the ratio between a degree and a radian, but I already knew this. My question was why radians are special and give the answer of 1 (which you skip entirely when you say, "We know that…"). Sorry if my question was unclear. $\endgroup$ – Sophie Alpert Feb 13 '11 at 18:48
  • $\begingroup$ @Ben: I have attempted a geometric interpretation. Kindly let me know what you think. $\endgroup$ – user17762 Feb 13 '11 at 22:33
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    $\begingroup$ $\lim_{x^\circ \rightarrow 0} \frac{\sin(x^\circ )}{x^\circ}$ is most certainly $1$, by any standard. Whatever your interpretation of $x^\circ$ is, it must be as a real variable, and $\lim_{y\rightarrow 0} \frac{\sin y}{y}=1$ for any real variable $y$. Typically (e.g. by Mathematica), $x^\circ$ is interpreted via $x^\circ=x\times\frac{\pi}{180}$. Perhaps you meant to write $\lim_{x\rightarrow 0} \frac{\sin (x^\circ )}{x} = \frac{\pi}{180}$ or $\lim_{x^\circ\rightarrow 0} \frac{\sin (x^\circ )}{x} = \frac{\pi}{180}$? Either would be correct. What you have is not. $\endgroup$ – Marcel Besixdouze Nov 16 '14 at 4:48

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.


Based on your comments in response to answers, you seem to be asking rather why radians are "special" in giving the answer $1$.

This really comes down to why is it that when we use radians, we have $$\lim_{x\to 0}\frac{\sin x}{x} = 1$$ because it comes down to why $(\cos x)' = -\sin x$ and not some other constant multiple of $\sin x$.

Think about the unit circle, parametrized by \begin{align*} x&=\cos t,\\ y&=\sin t \end{align*} (in whatever units you want to pick for $t$: radians, gradians, degrees, anything, doesn't matter here).

Whenever you parametrize a curve, you can change the parameter so that you traverse the curve slower, or faster, or in a different direction, or in a different way; here, if you replace $t$ with $-t$, you reparametrize the unit circle so that it is traversed clockwise, instead of counterclockwise. If you change $t$ to $2t$, you traverse it twice as fast but in the same direction; if you replace $t$ with $t+17$, then you start somewhere else; etc.

There is, however, what one might call a "normalized" parametrization, which normalizes the speed at which we traverse the curve. It is called the "arc-length parametrization", and the objective is to choose the parameter $t$ so that, if $t$ varies from $a$ to $b$, then the length of curve that is traversed is exactly $b-a$; that is, it takes exactly one unit of "time" to traverse one unit of "distance".

The arc-length parametrization has a number of advantages. For example, if you compute the tangent vector by taking $(x'(t),y'(t))$, in the arc length parametrization of a curve the tangent vector always has unit length. It also allows you to find the curvature of the curve much easier than with a run-of-the-mill parametrization.

Well, what is the arc-length parametrization of the unit circle? It is exactly the one in which it takes from $t=0$ to $t=2\pi$ to traverse the entire circle once; it is precisely the one in which an angle measured by $t$ cuts a circle arc of length $t$. It is, in short, measuring $t$ in radians. Radians give the arclength parametrization of the curve, which is why there are no pesky constants of proportionality when we take derivatives of the coordinate functions $\sin t$ and $\cos t$.

Now, no matter how you measure your angles, there will be some value of $b$ such that $$\int_0^b \sin_x t\,dt = 1.$$ (where by $\sin_x t$ I mean "sine, measuring the argument in $x$s"). The only question is what is the value of $b$; it exists for degrees, for radians, for gradians, for half-degrees, anything, really. Why does it turn out to be exactly $\pi/2$ for radians? Because when we do radians, which give the arc length parametrization, the lack of proportionality constants yield that the integral of the sine is the negative of the total change in the $x$ coordinate (under other parametrizations, it would be some constant multiple of this). If you start at $t=0$, the point $(1,0)$, when has the $x$-coordinate changed by exactly one unit? When we get to $(0,1)$, that is, when $t=\pi/2$.


This answer is similar to the answer of Marvis, but I think that it is conceptually a bit cleaner.

The $\sin(\theta)$ is defined as the y-coordinate of the point on the unit circle obtained by starting at the origin and moving a distance of $\theta$ along the unit circle in the counterclockwise direction.

So lets investigate what $\displaystyle \int_0^{\pi/2}\sin(\theta)$ would look like geometrically.

We are taking the interval $[0,\frac{\pi}{2}]$ and partitioning it into a million tiny pieces. On the unit circle, we have just broken the quarter circle in the first quadrant up into a million pieces. Then we need to multiply the lengths of each of these little segments by the height of the left endpoint of that line segment, and sum all of these values together. We need a geometric understanding of what that product looks like, i.e. how can we interpret $\sin(\theta)\Delta\theta$ geometrically?

enter image description here

Since we are looking at such a short segment of the circle, $\Delta\theta$ looks a lot a segment of the tangent line to the circle at that point. By geometry, the tangent line to the circle is perpendicular to the radius, and so the small triangle in the picture is similar to the large one. This lets us show that $\sin(\theta)\Delta\theta$ is approximately the small horizontal change in the picture (this is in fact a small change in cosine(\theta)).

Thus we are really just accumulating small horizontal distances. The total horizontal distance that you move from $\theta = 0$ to $\theta =\frac{\pi}{2}$ is 1, and so that is the integral.

If you have a solid geometric understanding of the fundamental theorem of calculus, and a solid geometric understanding of why the derivative of $\sin$ is $\cos$, then you will recognize that the proof above is really just weaving those two narratives together.

See also https://mathoverflow.net/questions/8846/proofs-without-words/8932#8932


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