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Reduction formula question

$I_{n}= \int_{0}^{\frac{\pi}{2}} \frac{\sin2nx}{\sin x} \text{d}x$, then show that, $I_{n}-I_{n-1}=(-1)^{n-1}\frac{2}{2n-1}$

How to begin for this question. I tried using by-parts, setting $u=\sin2nx$, and $v=\frac{1}{\sin x}$, as well as the vice versa, but I haven't made any progress.

What should I start to do here?

Anne Bauval
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lightningjay
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2 Answers2

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\begin{align*} I_n-I_{n-1}&=\int_0^{\fracπ2}\frac{\sin2nx-\sin2(n-1)x}{\sin x}dx =\int_0^{\fracπ2}\frac{2\sin x\cos(2n-1)x}{\sin x}dx\\ &=\frac2{2n-1}\bigg|\sin(2n-1)x\bigg|_0^{\fracπ2}=(-1)^{n-1}\frac2{2n-1}. \end{align*}

Gary
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aarbee
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I used this formula in this solution

I used $$\frac{2}{\pi}\int_{0}^{\frac{\pi}{2}}\frac{\sin\left(\left(2x-1\right)t\right)}{\sin\left(t\right)}\mathrm{d}t=\frac{\sin\left(\pi x\right)}{\pi}\left(\psi\left(1-\frac{x}{2}\right)-\psi\left(\frac{1-x}{2}\right)\right)-1$$

So in your case

$$\int_{0}^{\frac{\pi}{2}}\frac{\sin\left(2tx\right)}{\sin\left(t\right)}dt=\frac{\cos\left(\pi x\right)}{2}\left(\psi\left(\frac{3-2x}{4}\right)-\psi\left(\frac{1-2x}{4}\right)\right)-\frac{\pi}{2}$$

Math Attack
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  • Why would you go to such lengths when a simple trigonometric identity gives a direct method? – Mittens Feb 01 '25 at 13:54
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    @Mittens This answer also contains the case in which $x\in \mathbb{N}$ and in general in the problem it is not specified that $n\in \mathbb{N}$ (more than anything the integral is trivial for $n\in \mathbb{N}$, in fact it already has 1 vote for the question to be closed: I just tried to make it more interesting). – Math Attack Feb 01 '25 at 15:44
  • @Mittens it's just an alternative answer. And it doesn't even seem that "hyper" complicated to me, honestly. Indeed, I would appreciate the fact of dealing with the same problem with different approaches – Math Attack Feb 01 '25 at 18:16