Then, \( s = t - u \) and \( ds = -du \). The limits of integration change as follows: when \( s = 0 \), \( u = t \) and when \( s = t \), \( u = 0 \). Thus, we can rewrite the integral:
\[
I(t) = \int_t^0 e^{t-u} \sin(u) (-du) = \int_0^t e^{t-u} \sin(u) \,
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