A colleague of mine and I, in the course of teaching integral calculus for the umpteenth time, were wondering if we could expand the class of examples that our students are exposed to when computing Riemann integrals from the definition. Most such examples involve polynomial functions, and they work nicely because we have some well-known formulas, like

$\displaystyle\sum_{i=1}^n 1 = n$,

$\displaystyle\sum_{i=1}^n i = \frac{n(n+1)}{2}$,

and $\displaystyle\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$. (And certainly, there are others.)

But we wanted to expand the class of examples beyond polynomial functions, and started to think about trig functions. We came up with a computation for $\int_0^\pi \sin(x)\;dx$ using the definition of the Reimann integral, which I provide below.

Question: Is the computation below in the literature somewhere? If so, where?


Using a Riemann sum with right endpoints, we have

$$\int_{0}^\pi \sin(x)\; dx = \lim_{n \to \infty}\sum_{i=1}^n \sin(x_i) \Delta x$$ where $\Delta x = \frac{\pi}{n}$ and $x_i = i\Delta x$.

The trick is to rewrite the Riemann sum $\sum_{i=1}^n \sin(x_i) \Delta x$ as a telescoping sum using the difference of cosines identity: $$\cos(b) - \cos(a) = 2 \sin\left(\frac{a+b}{2}\right)\sin\left(\frac{a-b}{2}\right)$$

Setting $a = \frac{2i+1}{2}\Delta x$ and $b = \frac{2i-1}{2}\Delta x$ yields

$$\cos\left(\frac{2i-1}{2}\Delta x\right) - \cos\left(\frac{2i+1}{2}\Delta x\right) = 2 \sin\left(i \Delta x\right) \sin\left(\frac{\Delta x}{2}\right)$$

Solving for $\sin(i \Delta x)$ gives

$$\sin(i \Delta x) = \frac{\cos\left(\frac{2i-1}{2}\Delta x\right) - \cos\left(\frac{2i+1}{2}\Delta x\right)}{2 \sin\left(\frac{\Delta x}{2}\right)}$$

The Riemann sum now is $$\sum_{i=1}^n \sin(i \Delta x) \Delta x = \frac{\frac{\Delta x}{2}}{\sin\left(\frac{\Delta x}{2}\right)} \sum_{i=1}^n \left[\cos\left(\frac{2i-1}{2}\Delta x\right) - \cos\left(\frac{2i+1}{2}\Delta x\right)\right]$$

The latter sum is telescoping, and so $$\sum_{i=1}^n \sin(i \Delta x) \Delta x = \frac{\frac{\Delta x}{2}}{\sin\left(\frac{\Delta x}{2}\right)}\left[ \cos\left(\frac{\Delta x}{2}\right) - \cos\left(\frac{2n+1}{2}\Delta x\right)\right]$$

Now recalling that $\displaystyle \Delta x = \frac{\pi}{n}$ and taking $n \to \infty$ we have

$\begin{align} \int_0^\pi \sin(x)\;dx &= \lim_{n \to \infty} \frac{\frac{\Delta x}{2}}{\sin\left(\frac{\Delta x}{2}\right)}\left[ \cos\left(\frac{\Delta x}{2}\right) - \cos\left(\frac{2n+1}{2}\Delta x\right)\right]\\ & = 1 \cdot \left[ \cos(0)-\cos(\pi)\right]\\ & = 2 \end{align}$

One nice thing about this computation is that you can replace $\pi$ with an arbitrary upper limit of integration $c$ and compute that $\int_0^c \sin(x) \;dx = 1 - \cos(c)$. From there, one can easily compute $$\int_a^b \sin(x) \; dx = \int_0^b \sin(x) \; dx - \int_0^a \sin(x) \; dx = -\cos(b) + \cos(a)$$ (as predicted by FTC).

Question: Is the computation above in the literature somewhere? If so, where?

  • $\begingroup$ I don't know, but our teacher made us do this. $\int_a^b \sin(x)\,dx$ from reimann sum I mean. $\endgroup$
    – Guy
    Apr 3, 2014 at 18:43
  • $\begingroup$ This answer contains telescoping ideas. $\endgroup$
    – Ian Mateus
    Apr 3, 2014 at 18:43
  • $\begingroup$ I haven't seen it in a textbook or anywhere else, but I did it once myself in order to see how the limit $\sin x/x \rightarrow 1$ would play into the calculation. $\endgroup$ Apr 3, 2014 at 18:49
  • $\begingroup$ Not exactly "literature", but the sum $\sum \sin(n)$ is telescoped using a similar trick in this answer in order to apply the Dirichlet test to the series $\sum \frac{\sin(n)}{n}$. $\endgroup$
    – Mike F
    Apr 4, 2014 at 16:34

3 Answers 3


Below are some references I found in my "math library" at home this morning. Before posting just now, I quickly searched to see which were online and I've included links for those I found. The Boyer paper URL is a JSTOR item I don't have access to, but I've included the URL anyway because many here are probably at colleges/universities that have JSTOR access.

At the risk of pointing out the obvious, this method doesn't prove the Riemann integrability of ${\sin x},$ since equal length partitions are used along with endpoint evaluations. For instance, under this restriction the limit of the Riemann sums of the characteristic function of the rationals on a given compact interval exists. What this calculation does is give the value of the (Reimann) integral on a given compact interval under the assumption that the integral exists.

Direct quotes from original sources are indicated by italics. Within such a quote: (1) italics in the original is indicated by non-italics here; (2) brief additions and/or words of further explanation by me are indicated by non-italics between square braces "[stuff by me]"; (3) omissions in the original are indicated using "$[\dots]$".

[1] [Author not known], Sur la sinussoide [On the sinusoid], Question D'Examen, Nouvelles Annales de Mathématiques (1) 7 (1848), 436-437.

The (net) area above the $x$-axis and below the graph of $y = \sin{x},$ between $x = x_1$ and $x = x_2$ is found. The interval $[x_1, \, x_2]$ is divided into $n+1$ many abutting intervals each of length $h,$ so that $x_2 - x_1 = (n+1)h,$ and the identity $$h [\sin{x_1} + \sin{(x_1 + h)} + \sin{(x_2 + 2h)} + \cdots + \sin{(x_1 + nh)} \;\; = \;\; \frac{h \sin \left(x_1 + \frac{nh}{2}\right) \sin \frac{h}{2}(n+1)}{\sin{\frac{h}{2}}}$$ along with $\frac{nh}{2} = \frac{x_2 - x_1 - h}{2}$ is used to obtain the value $2\sin{\left(\frac{x_2 + x_1}{2}\right)} \sin{\left(\frac{x_2 - x_1}{2}\right)},$ which equals $\cos{x_1} - \cos{x_2}.$ Note: There is a typo at one point in the original, in which $\sin{(x_2 + h)}$ appears instead of $\sin{(x_2 + 2h)}.$ Also, the text only observes that the area's expression in terms of the sine function reduces to $1 - \cos{x_2}$ when $x_1 = 0.$

[2] William Elwood Byerly, Elements of the Integral Calculus, 2nd edition, Ginn and Company, 1892, xvi + 339 + 11 + 32 pages.

See Chapter VIII, Article 81, p. 76, Examples (2): By the aid of the trigonometrical formulas $$\cos{\theta} + \cos{2\theta} + \cos{3\theta} + \cdots + \cos{(n-1)\theta} = \frac{1}{2}\left[\sin{n\theta} \cot{\frac{\theta}{2}} - 1 - \cos{n\theta}\right],$$ [and] $$\sin{\theta} + \sin{2\theta} + \sin{3\theta} + \cdots + \sin{(n-1)\theta} = \frac{1}{2}\left[(1 - \cos{n\theta}) \cot{\frac{\theta}{2}} - \sin{n\theta}\right],$$ prove that $\;\;\int_{a}^{b}\cos{x}.dx = \sin{b} - \sin{a},\;$ and $\;\int_{a}^{b}\sin{x}.dx = \cos{a} - \cos{b}.$

[3] Carl Benjamin Boyer, History of the derivative and integral of the sine, Mathematics Teacher 40 #6 (October 1947), 267-275.

[4] Mark Bridger, A note on areas under a sine curve, The Pentagon 18 #1 (Fall 1958), 24-26.

At the time Mark Bridger was a student at the Bronx High School of Science (New York).

[5] Richard Courant and Fritz John, Introduction to Calculus and Analysis, Volume I, John Wiley and Sons (Interscience Publishers), 1965, xxiv + 661 pages.

See Chapter 2.2.e (Integration of $\sin x$ and $\cos x$), p. 135.

[6] Godfrey Harold Hardy, A Course of Pure Mathematics, 10th edition, Cambridge University Press, 1952, xii + 509 pages.

See Chapter VII, Article 164, p. 320, item 3. Calculate $\int_{a}^{b}x^2 \, dx,$ $\int_{a}^{b} \cos{mx} \, dx$ and $\int_{a}^{b} \sin{mx} \, dx$ by the method of Ex. 1. Note: Hardy's reference to "the method of Ex. 1" is simply the method of partitioning the interval $[a,b]$ into finitely many equal length subintervals, forming the relevant Riemann sum, and taking a limit. The needed trigonometric identities are not given as a hint, but I do notice that a closely related identity appears in Exercise 4 at the top of p. 323 for another purpose.

[7] Kenneth Sielke Miller and John Breffni Walsh, Elementary and Advanced Trigonometry, Harper and Brothers, 1962, xii + 350 pages.

See Chapter 10.3 (Solution of the Area Problem), pp. 215-217: (begins) We now turn to the second problem mentioned at the beginning of the chapter, namely, that of finding the area under one arch of a sine curve.

[8] Isidor Pavlovich Natanson, Summation of Infinitely Small Quantities, Topics in Mathematics, D. C. Heath and Company, 1963, viii + 59 pages.

See Chapter 6 (The Sinusoid), Articles 26-35, pp. 46-57. Specifically, see Article 29 (A trigonometric sum), Article 30 (A subsidiary inequality), Article 31 (The sine of an infinitely small angle), and Article 32 (The quadrature of the sinusoid) on pp. 46-51. The remaining articles in this chapter involve applications, mainly in calculating the volume of a rotating sinusoid and calculating effective (electrical) current.

[9] John Meigs Hubbell Olmsted, Intermediate Analysis. An Introduction to the Theory of Functions of One Real Variable, The Appleton-Century Mathematics Series, Appleton-Century-Crofts, 1956, xiv + 306 pages.

See Chapter 5, Article 503, p. 145, Exercise #19 & 20. The reader is asked to show that $\int_{a}^{b}\sin x \, dx = \cos{a} - \cos{b}$ and $\int_{a}^{b}\cos x \, dx = \sin{b} - \sin{a}$ by using equal length partitions of $[a,b].$ The relevant trig. identities are indirectly provided (by means of a hint, the reader is led to obtain these identities).

[10] Michael David Spivak, Calculus, 3rd edition, Publish or Perish, 1994, xiv + 670 pages.

See Chapter 15, p. 320, Problem 33.

[11] Isaac Todhunter, A Treatise on the Integral Calculus, 6th edition, MacMillan and Company, 1880, viii + 408 pages.

See Chapter IV, Article 38, pp. 52-53: (begins) Suppose we wish to find the integral of $\sin x$ between limits $a$ and $b$ immediately from the definition. By Art. 4 we have to find the limit $[\ldots]$ Note: The copy I found online is dated 1863, but what I've quoted seems to be the same calculation and it's in the same location in the book.

[12] Edwin Bidwell Wilson, Advanced Calculus, Ginn and Company, 1911, x + 566 pages.

See Chapter I, p. 30, Exercise 9. With the aid of the trigonometric formulas $[\ldots]$ show $(\alpha) \; \int_{a}^{b} \cos{x}\,dx = \sin{b} - \sin{a},\;\;$ [and] $\;\;(\beta) \; \int_{a}^{b} \sin{x}\,dx = \cos{a} - \cos{b}.$

(ADDED 3 DAYS LATER) I thought it would be useful to archive in my answer some references for evaluating definite integrals of other functions by Reimann sums. However, because I want to get this finished during my lunch hour and I don't want to substantially increase the length of this already long answer, I will use briefer citations.

Harding/Barnett, Solution to Calculus Problem #369, Amer. Math. Monthly 22 #6 (June 1915), 208-210.

Two solutions for the evaluation of $\int_{a}^{b} \ln{x}\,dx$. The second solution makes use of Fermat's method of geometric subdivision of the interval $[a,b].$

Barnett, Solution to Calculus Problem #366, Amer. Math. Monthly 22 #5 (May 1915), 168-169.

One solution for the evaluation of $\int_{a}^{b}{\sin}^{-1}x\,dx.$

Eli Maor, On the direct integration $\int_{a}^{b}x^{n}\,dx,$ Int. J. Math. Educ. Sci. Technol. 5 (1974), 199-200.

The method involves working with the sum of the left endpoint sums and the right endpoint sums. The method might be similar to what Otto Dunkel does in Note on the quadrature of the parabola, Amer. Math. Monthly 27 #3 (March 1920), 116-117.

Finally, the following are similar to the kind of things found in Natanson's Summation of Infinitely Small Quantities:

W. R. Longley, Some limit proofs in solid geometry, Amer. Math. Monthly 31 #4 (April 1924), 196-202.

Joseph B. Reynolds, Some applications of algebra to theorems in solid geometry, Mathematics Teacher 18 #1 (January 1925), 1-9.

Jos. B. Reynolds, Finding plane areas by algebra, Mathematics Teacher 21 #4 (April 1928), 197-203.

  • $\begingroup$ This is a great list of references! Thanks! $\endgroup$
    – wckronholm
    Apr 4, 2014 at 14:39

I suspect Tom Apostol's calculus textbook has this (i.e., finds $\int_0^\pi\sin x\,dx$ by using limits of Riemann sums without the fundamental theorem).

I question the place of Riemann sums in the curriculum. Rigorous definitions are unsuitable for a calculus-for-liberal-education course unless the students are unusual. Such a course should acquaint students with the reason why calculus is important in the course of human events over recent and coming centuries, and with the fact that it overcame difficulties like how to define $\text{rate} = \dfrac{\text{distance}}{\text{time}}$ when the distance and the time are both $0$, and how to to find $$\sum\text{force}\times\text{distance}$$ when there are infinitely many infinitely small distances, each with its own value of "force". The conventional calculus course is a watered-down version of a course for students who come in with a prior desire or a pre-identified need to understand calculus, rather than for students who are there in order to pay a price in homework for a grade to impress employers, and whom one should be trying to seduce into another course of action.

If you expect students to marvel at the fact that it's possible to find this integral by using limits of Riemann sums, I expect the way they will think of it and remember it is "We did some technical stuff and turned it in and got graded", and if you have them plod through finding the integral by antidifferentiating and substituting endpoint values, then the way they will think of it and remember it is "We did some technical stuff and turned it in and got graded", and they won't know the difference. (I'm hoping for a malpractice suit against every university that knowingly encourages unqualified students to take calculus, to the point where those are 99.9% of the ones who show up. They bring in tuition money.)

Another occasion for Riemann sums is numerical integration, but that doesn't seem to be your purpose.

  • 1
    $\begingroup$ Apostol's book gives the formulas for $\int_a^b \sin(x)\;dx$ and $\int_a^b \cos(x)\;dx$ followed by the statement "we shall not prove these formulas at this stage because they will be derived by an easier method in Chapter 2, with the help of differential calculus." $\endgroup$
    – wckronholm
    Apr 3, 2014 at 18:55
  • $\begingroup$ @wckronholm : OK. Apostol's book may still be a place to look if anyone wants similar material. Also, maybe Otto Toeplitz' "genetic approach" calculus book. $\endgroup$ Apr 3, 2014 at 18:58
  • $\begingroup$ $99.9\%$ may be unduly pessimistic. Perhaps I will be accused of being a rosy-eyed optimist, but $95\%$ seems more reasonable. $\endgroup$ Apr 3, 2014 at 19:15
  • $\begingroup$ @AndréNicolas : Depends on which institution. There are colleges that think of themselves as selective, at which the students are hard-working and intelligent, but have attitudes about math courses: They exist in order to provide an opportunity to show that they can follow instruction and work hard and get a grade, and they've always gotten "A"s in math, and they demand their right to get a course consisting of algorithms for them to follow to get an "A". $\endgroup$ Apr 3, 2014 at 19:36
  • $\begingroup$ @MichaelHardy I actually had my class run through this computation in small groups in class today. While it's true that many of them were longing for the Fundamental Theorem of Calculus they "learned" in high school and could give them the answer in about 5 seconds, there were more than a few who really were appreciating all of the pieces that went into the computation. In general, I think students benefit from exposure to this sort of activity because of the potential to broaden their reasoning skills. $\endgroup$
    – wckronholm
    Apr 3, 2014 at 20:15

Alternatively, you can use $e^{i \theta} = \cos(\theta) + i \sin(\theta)$ and the partial sum formula for Geometric series. Using left endpoint approximation instead, the sum we want to evaluate is $$ \sum_{k=0}^{n-1} \sin(k \, \Delta x)\,\Delta x$$ where $\Delta x = \frac{\pi}{n}$. Note that $\sum_{k=0}^{n-1} \sin(k \,\Delta x)$ is the imaginary part of $\sum_{k=0}^{n-1} e^{i k \,\Delta x}$, that is $\sum_{k=0}^{n-1} z^k$ where $z = e^{i \,\Delta x}$. This last sum has the closed form $\frac{1-z^n}{1-z} = \frac{2}{1-z}$ using $z^n = e^{i \pi} = -1$. We get $$ \sum_{k=0}^{n-1} \sin(k \,\Delta x)\,\Delta x =\mathrm{Im}\left( \frac{2}{1-e^{i\, \Delta x}} \right) \,\Delta x = \mathrm{Im}\left( \frac{2\,\Delta x}{1-e^{i \,\Delta x}} \right).$$ Finally, letting $n \to \infty$ so that $\Delta x \to 0$ we get $$\lim_{\Delta x \to 0} \frac{2\,\Delta x}{1-e^{i \,\Delta x}} = -2\left( \lim_{\Delta x \to 0 } \frac{e^{i \,\Delta x} - 1}{\Delta x} \right)^{-1} = -2\left( \frac{d}{dt} e^{it} \big|_{t=0} \right)^{-1} = -2i^{-1} = 2i$$ so that $$ \lim_{n \to \infty} \sum_{k=0}^{n-1} \sin(k \,\Delta x)\,\Delta x = \operatorname{Im}(2i) = 2.$$

  • $\begingroup$ Thanks. We were aware of this method already, and wanted to find an approach which did not rely on complex numbers. $\endgroup$
    – wckronholm
    Apr 4, 2014 at 16:05
  • $\begingroup$ @wckronholm: That makes sense, just thought I'd add it. $\endgroup$
    – Mike F
    Apr 4, 2014 at 16:28
  • $\begingroup$ @wckronholm: It has probably occurred to you already, but if you don't want to use complex numbers, you can still do integrals like $\int_0^1 e^x \ dx$ using geometric series. $\endgroup$
    – Mike F
    Apr 4, 2014 at 16:29
  • $\begingroup$ Thanks, that has indeed occurred to me, and I will be showing this to my students in about 20 minutes! $\endgroup$
    – wckronholm
    Apr 4, 2014 at 16:39

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