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Question Number 94123 by  M±th+et+s last updated on 17/May/20

$$provethat$$$${\int }_0^{\mathrm{\infty }}\frac{{sin}^{2n}\left(x\right)}{x^2}d={\int }_0^{\mathrm{\infty }}\frac{{sin}^{2n-1}\left(x\right)}{x}dx$$$$$$$$$$

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