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Dec 6, 2020 at 15:38 comment added HiRod Thank you very much!
Dec 6, 2020 at 15:36 vote accept HiRod
Dec 5, 2020 at 21:59 comment added Joe Silverman Might it be easier to rewrite your integrand as $$\frac{\sin(x)\cdot\sin^2(2x)}{2x^2}.$$ Or alternatively maybe $$\frac{2\sin^3(x)}{x^2}-\frac{2\sin^5(x)}{x^2}.$$
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