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I've stumbled across this integral: $$\int_{0}^{\pi/2} \left(\frac{\sin5x}{\sin x}\right)^2 \,dx $$

I was on a time limit and my intuition told me: $$\int_{0}^{\pi/2} \left(\frac{\sin5x}{\sin x}\right)^2 \,dx = \int_{0}^{\pi/2}5dx= 5x\Big|_0^{\pi/2}=\frac{5\pi}{2}$$

which turned out to be the right answer.

I'd like to know the correct way of integrating such a function. I've tried several techniques and all of them failed (just started integrals in my calc 1 class).

And is it just a coincidence that the first equality sign above hold or is there some explanation to it?

Quanto
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Avgustine
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  • Try the quintuple Angle formula, and you will find that you only need to compute the integral of one of the terms, the other terms because the periodicity is zero. – Always Mar 02 '24 at 15:49

1 Answers1

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Per symmetry $$I=\int_{0}^{\pi/2} \left(\frac{\sin5x}{\sin x}\right)^2 \,dx =\frac14 \int_{0}^{2\pi} \left(\frac{\sin5x}{\sin x}\right)^2 \,dx $$ With $\frac{\sin5x}{\sin x}=\frac{e^{i5x}- e^{-i5x}}{e^{ix}-e^{-ix}}=\sum_{k=-4}^4 e^{i k x} $ $$I= \frac14 \sum_{k,j=-4}^4\int_0^{2\pi}e^{i(k+j)x}dx=\frac14\int_0^{2\pi}5\ dx= \frac{5\pi}2 $$ where only the five pairs of $k+j=0$ survive the integration over $(0,2\pi)$ due to periodicity.

Quanto
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