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

1. Aug 25, 2008

### lkj-17

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

Anybody can give a hint how to do the integration.

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

2. Aug 26, 2008

### Ben Niehoff

The first integral can be easily done with integration by parts. For the second integral, use the sum-of-angles formula and then integrate by parts.