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Is it possible to do integration with degrees rather than radians with trigonometric functions?

For example, if I want to find $\int_{0}^{180} \sin(x) \,dx $ where $x$ is measured in degrees, I can write it as $\int_{0}^{180} \sin(\frac{x\pi}{180})\,dx $ where $x$ is measured in radians, however this gives me $\frac{360}{\pi}.$ And if I wanted to differentiate $\sin(x)$ where $x$ is in degrees, I can get $\frac{\pi}{180} \cos(x).$ To get the correct answer for the integral, which is $2$, I have to cancel $\frac{180}{\pi}$ down to 1; why?

ryang
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Nav Bhatthal
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  • Alright, thanks for the feedback. Say if my velocity was given by $f(t) = sin(t)$, with $t$ in degrees, would my displacement from $t=0$ and $t=180$ be $\frac{360}{\pi}$ or $2$? – Nav Bhatthal Jun 18 '23 at 16:00
  • Velocity is distance per unit time. To get displacement from velocity, you integrate over time. So when you say you want to integrate the velocity $f(t)=\sin(t)$ to get displacement, the $t$ in $\int_a^b\sin(t),\mathrm dt$ represents time. What do you mean by measuring the time $t$ in degrees? – David K Jun 19 '23 at 00:24
  • The short answer to your question is yes. It is done just about how you think it would be. See the answers for more explanation. What you are seeing, however, is that calculus furnishes one of the principle reasons we like to work with radians and not degrees. Calculus "knows" that the circumference of a unit circle is $2\pi$. – Charles Hudgins Jun 19 '23 at 15:05

4 Answers4

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For the integral $$ \int_0^\pi\sin(x)\,dx $$ you want to change variables according to $x = \frac\pi{180}x^\circ$. You can't just change $x$, you also have to change $dx$, which tells you "how big" a change in $x$ is. An increment of 1 degree is not the same as an increment of 1 radian, we need to have $dx = \frac\pi{180}dx^\circ$. This means $$ \int_0^\pi\sin(x)\,dx = \int_0^{180}\sin(\pi x^\circ/180)\frac\pi{180}dx^\circ = \frac\pi{180}\int_0^{180}\sin(\pi x^\circ/180)\,dx^\circ. $$ This is a different integral from what you wrote: $$ \int_0^{180}\sin(\pi x^\circ/180)\,dx^\circ = \frac{180}\pi\int_0^\pi\sin(x)\,dx. $$

Nicholas Todoroff
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I would propose an alternative way of removing the confusion here, which I learned from a mathematical physicist: Arguments to transcendental functions never ever have units. You must first convert the dimensionful quantity to a pure number, and only then can you evaluate functions on it. From this perspective, there is no "$x$ is in radians" or "$x$ is in degrees". The conventional $\sin$ function takes as input numbers (say, $\pi$) and returns numbers as output (say $\sin(\pi) = 0$).

If you want to interpret the "angle" $x$ as being measured by a different unit, then you are dealing with an entirely different function. Then there is no confusion. Let's call it $s_d(x)$ (d for degrees). So $s_d(0) = s_d(180) = 0$, $s_d(90) = 1$, etc. It is easy to see that for all $x$, $\sin(x) = s_d\left(\frac{180x}{\pi}\right)$. But now there is no way to be confused:

$$\int_{0}^{180} s_d(x)\,\mathrm{d}x = \frac{180}{\pi} \int_0^\pi \sin(y)\,\mathrm{d}y$$ $$\int_{0}^{\pi} \sin(x)\,\mathrm{d}x = 2$$ $$s_d'(x) = \frac{\pi}{180}c_d(x) \quad \text{(where $c_d$ is "cosine measured in degrees")}$$

There is no contradiction between the second identity and the familiar $\sin'(x) = \cos(x)$. $\sin$ and $s_d$ are different functions and you should not expect them to have the same derivative.

In the comments you ask:

Say if my velocity was given by $f(t) = \sin(t)$, with $t$ in degrees, would my displacement from $t=0$ and $t=180$ be $\frac{360}{\pi}$ or $2$?

I claim this is not a physically meaningful question. You say $t$ is measured in degrees, but you also say that your velocity is one-dimensional. Those two statements are inconsistent, because in one dimension there are no angles to speak of.

However, there is a physically meaningful situation that is close to what you said. Imagine a particle moving at a uniform speed around a circle, starting at the bottom and ending at the top. If the particle has unit speed, then the vertical component of the particle's velocity will be $\sin(t)$ whenever the angle of the particle from the (downward) vertical is $t$ radians. If the angle is in fact $D$ degrees, the vertical velocity will be $s_d(D)$. At every instant of the motion, if you have equal angles $D$ degrees and $t$ radians, both formulae will give the same answer.

Similarly, the vertical displacement of the particle at each instant will be $\cos(t)$ or $c_d(D)$, again depending on what system of units you choose. The final displacement is $2$ no matter what system you choose (as should be physically obvious).

But isn't there a contradiction there? Why do I differentiate to get the velocity when I measure in radians, but not when I measure in degrees? Why doesn't integrating the vertical velocity give me the displacement when I measure in degrees?

The answer is that you are not using the displacement formula correctly. The formula $v(t) = x'(t)$ presumes that the independent variable measures time in the system. But if your velocity is measured as a function of degrees, then you are not using absolute time as a parameter - you are using a scaled version of time, and the formula for calculating displacement from velocity is different in that case.

If you do choose time as your parameter, then you happen to get the same formula as if you measure in radians. This is one of the reasons physicists choose to use radians in their calculations.

(An interested reader may like to read the wikipedia article on Dimensional Analysis).

preferred_anon
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    To be more precise, it is not that arguments to functions never have units, because $x\mapsto2x$ and $x\mapsto x^2$ also are functions but their arguments often have units in physics problems. There are particular functions, such as $\exp$ and all functions derived from it (including the log, trig, and inverse trig functions), whose arguments must not have units. – David K Jun 19 '23 at 14:42
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    @DavidK You are right! I'll amend the answer to say "transcendental functions", but of course many polynomials are also excluded depending on whether the terms are commensurable. – preferred_anon Jun 19 '23 at 14:58
  • I agree with the opening paragraphs (I didn't read the rest): if the input but not the output of the natural sine function actually has units, then its function composition wouldn't make sense. Sure, we can allow units like in DavidK's "physicists' $x^2$ function", but such a concession is at best unnecessary. – ryang Jun 19 '23 at 15:01
  • @ryang It's quite important that dimensionful quantities can be multiplied (or you can't construct the dimensionless ratios that go into transcendental functions). I assume that is why David gave $x^2$ as an example. – preferred_anon Jun 19 '23 at 15:03
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The value of the integral you mention is indeed $\frac{360}{\pi}$, this was predictable, when you draw the graph of $f(x/\lambda)$ with $\lambda >0$, you dilate $f$ by $\lambda$, i.e. you get a wider function. For example you can draw both graphs on geogebra to see with your own eyes that the areas under the curves of $\sin$ and $\sin\left(\frac{180}{\pi} x\right)$ are completely different.
Same thing if you integrate, a $\lambda$ factor will ineluctably appear.

From now on, only work with radians, except when you speak with "civilians".

julio_es_sui_glace
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  • Alright, but there is another reason why I wanted to ask this, that just dawned on me. In my textbook there is some function $f(sin(x))$ where $x$ is explicitly measured in degrees. This is given as the velocity of a particle, and im tasked with finding the displacement (obviously the integral), but the answer to the problem is the "true" area, i.e in my case above the displacement would have been given as $2$, and not $\frac{360}{\pi}$. – Nav Bhatthal Jun 18 '23 at 15:52
  • Primitivation and differentiation are linear, so either you convert to radians at the beginning and then degree when you have done your computation or you just say that the derivative of $\sin$ is $\frac{\pi}{180} \cos$. I would do the first one if I were you, less confusion possible... – julio_es_sui_glace Jun 18 '23 at 15:59
  • Could you answer the question I have posted in my comment to ancient mathematician please? I would be very greatful. – Nav Bhatthal Jun 18 '23 at 16:12
  • @Mathguy If the problem you describe in your comment is the one that led to this question, then you should edit the question to show the statement of the problem exactly as it was shown to you with no words or symbols omitted. What you have described is not even a complete question, so let's see what the problem statement actually said and how we can make sense of it. – David K Jun 20 '23 at 13:23
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I have even asked such questions from my professor and he even warned me of a fact that the degree scale isn't that accurate, if we compare it to the radian scale. But, even if you want to continue with this, you must re-define "dx" as well...

Because the "dx" in radian scale is different from the "dx" in degree scale... Actually, 1 radian = $\frac{\pi}{180}$ degrees, and hence a degree is too small, if we compare it to a radian. So, its "dx" must also be as small as "dx" of radian scale.. So, we will multiply it with a factor of $\frac{\pi}{180}$ as well..

If I say dx for degree scale as dx', and dx of radian scale as dx, then it simply means that: $ dx = \frac{\pi}{180} dx'$

Hence, you need this replacement as well...

Now, your integral gets transformed like: $\int\sin({x})dx = \int\sin({x'}).\frac{\pi}{180}dx'$

Please note that I used x' for degrees and x for radians...

But if you take my advice, you must keep using radians only as they are more accurate way of measuring angles, and they're the true circular way of measurement.

Subrat Panda
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    I wonder what your professor means by the "fact that the degree scale isn't that accurate". I think of accuracy as a property of an approximation, whereas the conversion of radians to degrees is mathematically exact. What I find with the use of degrees in calculus is that things get much more complicated and confusing, for example how to integrate the sine function as exhibited in the question. So you might get results that are wildly incorrect for an application due to mistakes. Was that the kind of "accuracy" your professor was concerned about? – David K Jun 19 '23 at 00:17
  • Yeah, he said about accuracy just because if we use the degree scale in calculus, then it would really give wild results, and also, when we just convert from radian to degrees, then due to the irrational π, the result won't be accurate ofc.. Like if you treat π = 3.14, then you will say that 1 degree = 3.14/180 radians ~ but this is actually not accurate cuz π isn't truly equal to 3.14, it just goes on and on... We can just say that it's approximately same... This is what, he meant. – Subrat Panda Jun 19 '23 at 10:09
  • A right angle is $90$ degrees exactly; it is $\frac\pi2$ radians. So if you are worried about calculations involving $\pi$ being "inaccurate" because you can only practically write a $\pi$ to a finite number of digits, the degree measurement is more accurate in this frequently-encountered case. What makes angles in radians actually accurate is that when we get some multiple or fraction of $\pi$ radians we don't replace $\pi$ by $3.14$; we just write $\pi$. The correct response to the inaccuracy of "$1$ degree = $3.14/180$ radians" is: don't do that. – David K Jun 19 '23 at 13:34
  • Hmm I get it.. Then maybe, he would have just said it because the degree scale involves a lot of "extra" work, if we use it in calculus instead of using the radian scale... – Subrat Panda Jun 19 '23 at 15:48
  • @SubratPanda The point is that calling the degree scale "inaccurate" is wrong (and certainly inaccurate); what your professor was trying to say is that doing calculus in degrees mode is error-prone. (Sure, substituting 3.14/180 radians for 1 degree is imprecise, and claiming that they are exactly equal is inaccurate, but your professor was not trying to make this point at all.) – ryang Jun 21 '23 at 13:24
  • Yes, I get it... – Subrat Panda Jun 21 '23 at 16:15