@Kitonum Your (edited) approach is smooth, yes, but it computes the contribution from x=5..infinity as this:
int((sin(t)^2)^exp(t), t = 5 .. infinity, method = _d01amc, numeric);
But that looks quite wrong. (It's very close to 3.73051*10^-8 which wolframalpha.com gave me, btw, which I find funny.)
Consider just this portion,
int((sin(t)^2)^exp(t), t = 5 .. 20, numeric);
Shouldn't the integral for x=5..infinity be at least that much? It seems that the approach to force method=_d01amc neglects a significant portion of the integral from x=5..infinity.
int((sin(t)^2)^exp(t), t = 5 .. infinity, numeric);
I don't much like the look of that result 0.04401097083 for x=5..infinity, either, because it's the same as the result for x=5..20. I distrust it because of further missed portions like this:
evalf(Int(t->(sin(t)^2)^exp(t), 20.42 .. 20.425 ));
And so on, in ever-narrower bands around odd multiples of Pi/2. I haven't thought much about whether it converges. The total integral from x=5..infinity might be a great as 0.0441 or so.
What to do, then? If a function were devised to compute each contribution from a periodic subrange, then `evalf/Sum` might be able to tell whether their numeric sum converges quickly enough to validate. Umph.