Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

A book asks me to prove that:

$$\int_0^{\pi}\sin(x)\,dx = 2$$

Using the identity:

$$\sin\left(\frac{\pi}{n}\right) + \sin\left(\frac{2\pi}{n}\right) + \cdots + \sin\left(\frac{n\pi}{n}\right) = \frac{\cos\left(\frac{\pi}{2n}\right)-\cos\left(\frac{(2n+1)\pi}{2n}\right)}{2\sin\left(\frac{\pi}{2n}\right)}$$

And the famous $\lim_{x\to0}\frac{\sin(x)}{x} = 1$

What I tried:

Using the Right Riemann Sum Method:

$$\int_a^{b}\sin(x)\,dx \approx \Delta x\left[f(a + \Delta x) + f(a + 2\,\Delta x) + \cdots + f(b)\right]$$

By taking $\Delta x = \frac{\pi}{n}$, $a = 0$ and $b = \pi$ we have:

$$\int_0^{\pi}\sin(x)\,dx \approx \Delta x\left[f(\Delta x) + f(2\,\Delta x) + \cdots + f\left(\frac{n\pi}{n}\right)\right] = \frac{\pi}{n}\left[\sin\left(\frac{\pi}{n}\right) + \sin\left(\frac{2\pi}{n}\right) + \cdots + \sin\left(\frac{n\pi}{n}\right)\right] = \frac{\pi}{n}\left[\frac{\cos\left(\frac{\pi}{2n}\right)-\cos\left(\frac{(2n+1)\pi}{2n}\right)}{2\sin\left(\frac{\pi}{2n}\right)}\right]$$


$$\int_0^\pi \sin(x)\,dx = \lim_{n\to\infty} \frac{\pi}{n}\frac{\cos\left(\frac{\pi}{2n}\right)-\cos\left(\frac{(2n+1)\pi}{2n}\right)}{2\sin\left(\frac{\pi}{2n}\right)}$$

I can't see, however, how to prove this limit to be $=2$

share|cite|improve this question

3 Answers 3

up vote 3 down vote accepted

Using $$\cos A-\cos B=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}$$ with $$\lim_{n\to\infty}\frac{\pi/(2n)}{\sin(\pi/(2n))}=1,$$ We have $$\frac{\pi/(2n)}{\sin(\pi/(2n))}\cdot (-2)\sin\left\{\frac{1+(2/n)}{2}\pi\right\}\sin\left(-\frac{\pi}{2}\right)\to 1\cdot (-2)\cdot 1\cdot (-1)=2\ (n\to \infty)$$

share|cite|improve this answer
An elementar approach :) thank you – Guerlando OCs Aug 30 '14 at 0:17
@GuerlandoOCs: I agree with that Adriano's answer is better because we don't need to use anything:) – mathlove Aug 30 '14 at 0:18

Notice that we can rewrite the limit as: $$ \left[\lim_{n\to\infty} \frac{\frac{\pi}{2n}}{\sin\left(\frac{\pi}{2n}\right)} \right]\left[\cos\left(\lim_{n\to\infty}\frac{\pi}{2n}\right)-\cos\left(\lim_{n\to\infty}\frac{(2n+1)\pi}{2n}\right) \right] = 1 \cdot [\cos 0 - \cos \pi] = 2 $$

share|cite|improve this answer
Excelent! I'm in doubt in which answer I take. Yours is simpler but less 'elementar'. – Guerlando OCs Aug 30 '14 at 0:16
+1 nice! I don't need to use anything:) – mathlove Aug 30 '14 at 0:17
Less elementar, but much more mathematical beauty. +1. – user153012 Aug 30 '14 at 0:20
Do you guys have knowledge about when this integral was first discovered (not in the antidifferentiation way, but the elementar way). Was it before Newton? – Guerlando OCs Aug 30 '14 at 0:21
nice answer! it will be slightly better if you change $\cos\left(\lim_{n\to\infty}\frac{(2n+1)\pi}{2n}\right)$ to $\cos\left(\pi+\lim_{n\to\infty}\frac{\pi}{2n}\right)$ – mike Aug 30 '14 at 0:28

Without simplifying your last expression (thr numerator is twice the same term of it), we could use Taylor series and, for large values of $n$, you could write $$\cos\left(\frac{\pi}{2n}\right)=1-\frac{\pi ^2}{8 n^2}+\frac{\pi ^4}{384 n^4}+O\left(\left(\frac{1}{n}\right)^5\right)$$ $$\cos\left(\frac{(2n+1)\pi}{2n}\right)=-1+\frac{\pi ^2}{8 n^2}-\frac{\pi ^4}{384 n^4}+O\left(\left(\frac{1}{n}\right)^5\right)$$ $$\sin\left(\frac{\pi}{2n}\right)=\frac{\pi }{2 n}-\frac{\pi ^3}{48 n^3}+O\left(\left(\frac{1}{n}\right)^5\right)$$ So $$\frac{\pi}{n}\frac{\cos\left(\frac{\pi}{2n}\right)-\cos\left(\frac{(2n+1)\pi}{2n}\right)}{2\sin\left(\frac{\pi}{2n}\right)}=2-\frac{\pi ^2}{6 n^2}+O\left(\left(\frac{1}{n}\right)^4\right)$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.