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Skilled2022-01-19Added 437 answers
You can writesin(x)sin(x2)=12[cos(x2−x)−cos(x2+x)]and let u=x2−x,v=x2+x to obtain∫1tsin(x)sin(x2)dx=12∫1t[cos(x2−x)−cos(x2+x)] =14[∫0t2−tcos(u)u+14du−∫2t2+tcos(v)v+14dv]The convergence of this expression as t→∞ is ensured by Dirichlet's test for integrals or integration by parts.
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