To solve this problem, we need to find the Laplace transform of the integral:

$L\left\{ \int_0^t u \sin 2u \, du \right\}$

Solution Steps

1. Recognize the Laplace Transform Property: For an integral of the form $F(t) = \int_0^t f(u) \, du$, the Laplace transform $L\{F(t)\} = \frac{L\{f(t)\}}{s}$.

2. Set up the function $f(u) = u \sin 2u$: We need the Laplace transform of $u \sin 2u$, which we denote as $L\{u \sin 2u\}$.

3. Use the Laplace Transform of $t \sin(at)$: The Laplace transform $L\{t \sin(at)\} = \frac{2a s}{(s^2 + a^2)^2}$.

   Here, $a = 2$, so the transform of $u \sin 2u$ is: $L\{u \sin 2u\} = \frac{4s}{(s^2 + 4)^2}$

4. Apply the Integral Property: Since $F(t) = \int_0^t u \sin 2u \, du$, we divide by $s$: $L\left\{ \int_0^t u \sin 2u \, du \right\} = \frac{\frac{4s}{(s^2 + 4)^2}}{s} = \frac{4s}{s(s^2 + 4)^2}$

5. Simplify: This simplifies to: $\frac{4}{(s^2 + 4)^2}$

Conclusion

The correct answer is (c) $\frac{4}{(s^2 + 4)^2}$.

Do you want more details or have any questions?

Here are some related questions for further practice:

1. How would you find the Laplace transform of $\int_0^t t^2 \cos(3t) \, dt$?
2. What is the Laplace transform of $t^2 \sin(5t)$?
3. How can we use Laplace transforms to solve differential equations with integrals?
4. What is the Laplace transform of $\int_0^t e^{3u} \cos(u) \, du$?
5. How do you derive the Laplace transform formula for $t \sin(at)$?

Tip: Remember that the Laplace transform of an integral can be simplified by dividing by $s$ after transforming the inner function.