Evaluate Integral 0 to 2pi (3t+2)sin(t^3+2t+1)dt if possible

Answer 1

of course it's possible. Note that if
u = t^3+2t+1
du = 2t^2+2
and the integrand just becomes sin(u) du

Answer 2

except that the given contained (3t+2) instead of the expected (3t^2+2)

I suspect a misprint or typo

secondly if
u = t^3+2t+1
du = 3t^2+2