Evaluate ∫t0essin(t−s)ds by using Integration by parts.
If we let u=sin(t−s) and dv=esds, then
du=−cos(t−s)dsv=es
So,
∫t0essin(t−s)ds=uv−∫vdu=essin(t−s)−∫es(−cos(t−s))ds=essin(t−s)+∫escos(t−s)ds
To evaluate ∫escos(t−s)ds, we must use Integration by parts once more...
Hence, if we let u1=cos(t−s) and dv1=esds , thendu1=sin(t−s)dsv1=es
So, ∫escos(t−s)ds=u1v1−∫v1du1=escos(t−s)−∫essin(t−s)ds
Going back to the first equation,
∫t0essin(t−s)ds=essin(t−s)+[escos(t−s)−∫essin(t−s)ds]
Combining like terms
2∫t0essin(t−s)ds=essin(t−s)+escos(t−s)∫t0essin(t−s)ds=essin(t−s)+escos(t−s)2=es2[sin(t−s)+cos(t−s)]
Evaluating from 0 to t, we have
=et2−12sint−12cost=12(et−sint−cost)
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