Find the laplace transform of t e^t sin2t cost

To find the Laplace transform of the function \( f(t) = t e^t \sin(2t) \cos(t) \), we can proceed as follows:

### Step-by-Step Solution

1. **Simplify the Expression Using Trigonometric Identity:**
   Rewrite \( \sin(2t) \cos(t) \) using the identity:
   \[
   \sin(2t) \cos(t) = \frac{1}{2} \left( \sin(3t) + \sin(t) \right).
   \]
   Thus,
   \[
   f(t) = t e^t \left( \frac{1}{2} \sin(3t) + \frac{1}{2} \sin(t) \right).
   \]
   Simplifying, we get:
   \[
   f(t) = \frac{t e^t}{2} \sin(3t) + \frac{t e^t}{2} \sin(t).
   \]

2. **Use the Linearity of the Laplace Transform:**
   The Laplace transform \( \mathcal{L}\{f(t)\} \) can be split as:
   \[
   \mathcal{L}\{f(t)\} = \frac{1}{2} \mathcal{L}\{t e^t \sin(3t)\} + \frac{1}{2} \mathcal{L}\{t e^t \sin(t)\}.
   \]

3. **Apply the Laplace Transform Formula for \( t e^{at} \sin(bt) \):**
   The Laplace transform of \( t e^{at} \sin(bt) \) is given by:
   \[
   \mathcal{L}\{t e^{at} \sin(bt)\} = \frac{2b(s - a)}{((s - a)^2 + b^2)^2}.
   \]

   Now we apply this formula to each term:

   - **First Term:** For \( t e^t \sin(3t) \), we have \( a = 1 \) and \( b = 3 \):
     \[
     \mathcal{L}\{t e^t \sin(3t)\} = \frac{2 \cdot 3 \cdot (s - 1)}{\left((s - 1)^2 + 3^2\right)^2} = \frac{6(s - 1)}{((s - 1)^2 + 9)^2}.
     \]

   - **Second Term:** For \( t e^t \sin(t) \), we have \( a = 1 \) and \( b = 1 \):
     \[
     \mathcal{L}\{t e^t \sin(t)\} = \frac{2 \cdot 1 \cdot (s - 1)}{\left((s - 1)^2 + 1^2\right)^2} = \frac{2(s - 1)}{((s - 1)^2 + 1)^2}.
     \]

4. **Combine Results:**
   Substitute back into the expression for \( \mathcal{L}\{f(t)\} \):
   \[
   \mathcal{L}\{f(t)\} = \frac{1}{2} \cdot \frac{6(s - 1)}{((s - 1)^2 + 9)^2} + \frac{1}{2} \cdot \frac{2(s - 1)}{((s - 1)^2 + 1)^2}.
   \]
   Simplifying, we get:
   \[
   \mathcal{L}\{f(t)\} = \frac{3(s - 1)}{((s - 1)^2 + 9)^2} + \frac{(s - 1)}{((s - 1)^2 + 1)^2}.
   \]

### Final Answer
\[
\mathcal{L}\{t e^t \sin(2t) \cos(t)\} = \frac{3(s - 1)}{((s - 1)^2 + 9)^2} + \frac{(s - 1)}{((s - 1)^2 + 1)^2}.
\]

Would you like further explanation on any step or have any questions?

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Here are 5 related questions to deepen your understanding:

1. How would you find the Laplace transform of \( t e^t \cos(2t) \)?
2. What if the function was \( t^2 e^t \sin(2t) \cos(t) \)? How would that change the Laplace transform?
3. Can you use this technique to find the Laplace transform of \( t e^{2t} \sin(3t) \)?
4. How does the Laplace transform formula change if we have \( t^n e^{at} \sin(bt) \) instead?
5. What happens to the Laplace transform if we have only \( \sin(2t) \cos(t) \) without \( t e^t \)?

**Tip:** Always check if trigonometric identities can simplify expressions before applying transforms!

Find the Laplace transform of t e^t sin2t cost

Find the Laplace transform of the function t e^t sin(2t) cos(t) using trigonometric identities and Laplace formulas. Learn how to simplify the expression and apply the Laplace formula for functions of the form t e^{at} sin(bt). Suitable for college-level calculus and engineering students.

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