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Dividing given interval into n equal parts and for every interval taking ϵ as an midpoint of an interval find integral sum of a function and compute it.

$y=\sin x$ $x \in[0,\pi]$

$\triangle x_i= \frac{\pi}{n}i$

$\epsilon_i= \frac{0+\frac{\pi}{n}i+0+\frac{\pi}{n}(i+1)}{2} = \frac{2i+1}{2n}\pi$

$$\sum_{i=0}^{n-1}\sin\left(\frac{(2i+1)\pi}{2n}\right)\cdot\frac{\pi}{n} = \frac{\pi}{n}\sum_{i=0}^{n-1}\sin\left(\frac{(2i+1)\pi}{2n}\right)$$ from here I can't continue

unit 1991
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  • Does this answer your question? https://math.stackexchange.com/questions/2048590/how-do-we-know-what-the-integral-of-sin-x-is – Mark Sep 26 '21 at 12:04
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    @Mark No because at that question $\epsilon$ is chosen as a right endpoint we need midpoint. – unit 1991 Sep 26 '21 at 12:29
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    Since the function is integrable, you can choose any points. Not sure if you can easily compute it specifically with midpoints. – Mark Sep 26 '21 at 12:33
  • @Mark Yes I know my textbook says choose midpoints. – unit 1991 Sep 26 '21 at 12:34
  • This should answer your question, and has a wide variety of great proofs: https://math.stackexchange.com/questions/17966/how-can-we-sum-up-sin-and-cos-series-when-the-angles-are-in-arithmetic-pro – TomKern Sep 26 '21 at 13:34
  • @TomKern In that post I am getting $S \times \sin\frac{d}{2}$ = $\frac{1}{2}$ ($cos(a-\frac{d}{2}) - cos(a+\frac{d(2n+1)}{2})$ – unit 1991 Sep 26 '21 at 14:08
  • but it should be $\sin(n \times \frac{d}{2})\ \times \sin\biggl( \frac{2 a + (n-1)\cdot d}{2}\biggr)$ – unit 1991 Sep 26 '21 at 14:08

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