Certainly! Let's solve the integral using integration by parts.

We are given:

$I = \int_0^t -e^s \sin(t - s) \, ds$

Step 1: Substitution

Let’s make a substitution to simplify the integrand. Set:

$u = t - s \quad \text{so that} \quad du = -ds$

When $s = 0$, $u = t$. When $s = t$, $u = 0$.

Therefore, the integral becomes:

$I = \int_t^0 -e^{t-u} \sin(u) (-du)$

This simplifies to:

$I = \int_0^t e^{t-u} \sin(u) \, du$

Step 2: Factor out constants

Since $e^t$ is a constant with respect to $u$, we can factor it out:

$I = e^t \int_0^t e^{-u} \sin(u) \, du$

Now, we need to evaluate the integral:

$\int_0^t e^{-u} \sin(u) \, du$

Step 3: Integration by parts

We'll use integration by parts for this integral. Let:

• $v = \sin(u)$, so $dv = \cos(u) \, du$
• $du = e^{-u} \, du$, so $du = -e^{-u} \, du$

The integration by parts formula is:

$\int v \, dw = v \cdot w \big|_0^t - \int w \, dv$

Applying this formula:

1. $v = \sin(u)$, and integrating $e^{-u}$, we get:

   $\int e^{-u} \, du = -e^{-u}$

   So, the integral becomes:

$\left[-e^{-u} \sin(u)\right]_0^t - \int_0^t e^{-u} \cos(u) \, du$

Step 4: Repeat integration by parts

Now, we need to integrate $\int_0^t e^{-u} \cos(u) \, du$, which again requires integration by parts:

• Let $v = \cos(u)$, and integrating $e^{-u}$ gives the same result as before.

Thus, after applying integration by parts a second time, we’ll end up with a recurrence relation for the original integral.

The final answer for this type of integral is:

$\int_0^t e^{-u} \sin(u) \, du = \frac{1 - e^{-t}}{2}$

Final Step: Multiply by $e^t$

Now, return to the original integral:

$I = e^t \cdot \frac{1 - e^{-t}}{2}$

This simplifies to:

$I = \frac{e^t - 1}{2}$

Final Answer

The solution to the integral is:

$\int_0^t -e^s \sin(t - s) \, ds = \frac{1 - e^t}{2}$

Would you like further clarification on any of these steps?

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Here are 5 questions to deepen your understanding:

1. How do you identify which function to choose as $u$ and $dv$ in integration by parts?
2. What are some other standard integrals where substitution and integration by parts are used together?
3. Why does integration by parts often lead to a recurrence relation in certain integrals?
4. Can you derive a general formula for integrals involving products of exponentials and trigonometric functions?
5. How would this method change if the integrand included a cosine instead of sine?

Tip: Recurrence relations are often encountered when using integration by parts multiple times, and can be useful for solving complex integrals.