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

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

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

To evaluate the integral ∫[(3t+2)sin(t^3+2t+1)]dt over the interval [0, 2π], we can use integration by parts.

Let's denote u = (3t + 2) and dv = sin(t^3 + 2t + 1)dt. Then we can find du and v by differentiating and integrating respectively:

du = d/dt[(3t + 2)]dt = 3dt
v = ∫sin(t^3 + 2t + 1)dt

To find v, we can make a substitution:
Let's denote w = t^3 + 2t + 1. Then dw = (3t^2 + 2)dt, which gives us dt = dw / (3t^2 + 2).

Now, we can rewrite v in terms of w:
v = ∫sin(w) * (dw / (3t^2 + 2))

Next, we need to substitute the limits of integration:
When t = 0, w = (0)^3 + 2(0) + 1 = 1
When t = 2π, w = (2π)^3 + 2(2π) + 1

Now we can evaluate the integral by applying the integration by parts formula:
∫[(3t+2)sin(t^3+2t+1)]dt = (u * v) - ∫(v * du)

Plugging in the values, we have:
∫[(3t+2)sin(t^3+2t+1)]dt = [(3t + 2) * (-cos(w) / (3t^2 + 2))] - ∫[-cos(w) * (3dt / (3t^2 + 2))]

Simplifying further, we get:
∫[(3t+2)sin(t^3+2t+1)]dt = (-(3t + 2)cos(w) / (3t^2 + 2)) + ∫[cos(w) * (3dt / (3t^2 + 2))]

At this point, we can substitute back for t and simplify the integral.

However, it is important to note that the integral might not have a closed-form solution in terms of elementary functions. In such cases, the integral can be approximated using numerical methods, such as Simpson's rule or numerical integration techniques available in software tools.