Both Int {(t-tau)*sin(a*tau)}d(tau) and Int {(tau)*sin(a*(t-tau)}d(tau) will give the same answer (a*t-sin(a*t))/(a^2), where tau = 0..t(adsbygoogle = window.adsbygoogle || []).push({});

Anybody can give a hint how to do the integration.

Personally, I think neither integration by parts nor substitution are suitable methods.

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# Evaluate (t-tau)*sin(a*tau) with respect to tau, tau = 0t

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