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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:

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

So

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

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

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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 $$\frac{\pi/(2n)}{\sin(\pi/(2n))}\cdot (-2)\sin\left\{\frac{1+(2/n)}{2}\pi\right\}\sin\left(-\frac{\pi}{2}\right)\to 1\cdot (-2)\cdot 1\cdot (-1)=2\ (n\to \infty)$$

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

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

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Excelent! I'm in doubt in which answer I take. Yours is simpler but less 'elementar'. –  Guerlando OCs Aug 30 at 0:16
    
+1 nice! I don't need to use anything:) –  mathlove Aug 30 at 0:17
    
Less elementar, but much more mathematical beauty. +1. –  user153012 Aug 30 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 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 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 $$\frac{\pi}{n}\frac{\cos\left(\frac{\pi}{2n}\right)-\cos\left(\frac{(2n+1)\pi}{2n}\right)}{2\sin\left(\frac{\pi}{2n}\right)}=2-\frac{\pi ^2}{6 n^2}+O\left(\left(\frac{1}{n}\right)^4\right)$$

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