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?

---
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.

Learn how to solve the integral using integration by parts for -e^s * sin(t - s). This detailed solution covers substitution, factorization, integration by parts, and final simplification. Suitable for undergraduate students and anyone studying calculus and integration techniques.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation