definite integral of t^2sint(2t)dt with an upper limit of 8\pi and a lower limit of 0

To solve the definite integral \(\int_{0}^{8\pi} t^2 \sin(2t) \, dt\), we'll use integration by parts.

### Step 1: Identify \( u \) and \( dv \)
We can apply integration by parts twice, starting with:
\[
u = t^2 \quad \text{and} \quad dv = \sin(2t) \, dt
\]
Differentiate and integrate to find \(du\) and \(v\):
\[
du = 2t \, dt \quad \text{and} \quad v = \int \sin(2t) \, dt
\]
To integrate \( \sin(2t) \), recall:
\[
v = -\frac{1}{2} \cos(2t)
\]

### Step 2: Apply the integration by parts formula
Using the formula:
\[
\int u \, dv = uv - \int v \, du
\]
Substitute the values:
\[
\int t^2 \sin(2t) \, dt = -\frac{1}{2} t^2 \cos(2t) + \frac{1}{2} \int 2t \cos(2t) \, dt
\]
Simplify:
\[
\int t^2 \sin(2t) \, dt = -\frac{1}{2} t^2 \cos(2t) + \int t \cos(2t) \, dt
\]

### Step 3: Apply integration by parts again
For the remaining integral \(\int t \cos(2t) \, dt\), apply integration by parts again:
\[
u = t \quad \text{and} \quad dv = \cos(2t) \, dt
\]
Differentiate and integrate:
\[
du = dt \quad \text{and} \quad v = \frac{1}{2} \sin(2t)
\]
Substitute into the formula:
\[
\int t \cos(2t) \, dt = \frac{1}{2} t \sin(2t) - \frac{1}{2} \int \sin(2t) \, dt
\]
Integrate \(\sin(2t)\) again:
\[
\int \sin(2t) \, dt = -\frac{1}{2} \cos(2t)
\]
Thus:
\[
\int t \cos(2t) \, dt = \frac{1}{2} t \sin(2t) + \frac{1}{4} \cos(2t)
\]

### Step 4: Combine the results
Now combine all the pieces:
\[
\int t^2 \sin(2t) \, dt = -\frac{1}{2} t^2 \cos(2t) + \frac{1}{2} \left( \frac{1}{2} t \sin(2t) + \frac{1}{4} \cos(2t) \right)
\]
Simplify the expression:
\[
\int t^2 \sin(2t) \, dt = -\frac{1}{2} t^2 \cos(2t) + \frac{1}{4} t \sin(2t) + \frac{1}{8} \cos(2t)
\]

### Step 5: Evaluate the definite integral
Now evaluate the definite integral from 0 to \(8\pi\):
\[
\left[-\frac{1}{2} t^2 \cos(2t) + \frac{1}{4} t \sin(2t) + \frac{1}{8} \cos(2t)\right]_{0}^{8\pi}
\]
First, evaluate at the upper limit \( t = 8\pi \):
\[
-\frac{1}{2} (8\pi)^2 \cos(16\pi) + \frac{1}{4} (8\pi) \sin(16\pi) + \frac{1}{8} \cos(16\pi)
\]
Since \(\cos(16\pi) = 1\) and \(\sin(16\pi) = 0\):
\[
-\frac{1}{2} (8\pi)^2 (1) + 0 + \frac{1}{8}(1) = -32\pi^2 + \frac{1}{8}
\]
Next, evaluate at the lower limit \( t = 0 \):
\[
-\frac{1}{2} (0)^2 \cos(0) + \frac{1}{4} (0) \sin(0) + \frac{1}{8} \cos(0) = 0 + 0 + \frac{1}{8} = \frac{1}{8}
\]
Thus, the definite integral is:
\[
\left(-32\pi^2 + \frac{1}{8}\right) - \frac{1}{8} = -32\pi^2
\]

So, the result of the definite integral is:
\[
\boxed{-32\pi^2}
\]

Do you have any questions, or would you like further details? Here are some additional related questions:

1. How would the integral change if the limits were \(\pi\) and 0?
2. Can you compute the integral \(\int_{0}^{2\pi} t^2 \cos(2t) \, dt\)?
3. How does integration by parts apply to trigonometric integrals in general?
4. What are the applications of this integral in physics or engineering?
5. Can this integral be solved using any other methods like the tabular integration by parts?

**Tip:** When working with trigonometric integrals, always consider the periodicity of sine and cosine functions, as they can simplify the evaluation of definite integrals.

Learn how to compute the definite integral of t^2 sin(2t) from 0 to 8π using integration by parts. This problem involves advanced techniques in calculus and trigonometric integrals. Get step-by-step instructions and a detailed solution.

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