To solve the integral \( \int \sin(3 - t) \, dt \), let's proceed step by step:

### Step 1: Recognize the structure
The function inside the sine is \( 3 - t \), which is a linear expression in \( t \). This suggests a simple substitution might help.

### Step 2: Use substitution
Let:
\[
u = 3 - t
\]
Then, the derivative of \( u \) with respect to \( t \) is:
\[
\frac{du}{dt} = -1 \quad \Rightarrow \quad dt = -du
\]

### Step 3: Substitute in the integral
Substituting \( u = 3 - t \) and \( dt = -du \) into the original integral gives:
\[
\int \sin(3 - t) \, dt = \int \sin(u) (-du) = -\int \sin(u) \, du
\]

### Step 4: Solve the integral
We know that:
\[
\int \sin(u) \, du = -\cos(u)
\]
Therefore:
\[
-\int \sin(u) \, du = \cos(u)
\]

### Step 5: Substitute back for \( u \)
Since \( u = 3 - t \), the result becomes:
\[
\cos(3 - t) + C
\]
where \( C \) is the constant of integration.

### Final Answer:
\[
\int \sin(3 - t) \, dt = \cos(3 - t) + C
\]

Would you like more details or have any questions?

Here are 5 related questions to expand on this topic:
1. What is the integral of \( \cos(3 - t) \, dt \)?
2. How would you solve \( \int \sin(2t + 1) \, dt \)?
3. What substitution method is generally useful for solving integrals like \( \int \sin(ax + b) \, dx \)?
4. How does the result change if the integral had a cosine instead of a sine, i.e., \( \int \cos(3 - t) \, dt \)?
5. Can you apply integration by parts to \( \int t \sin(3 - t) \, dt \)?

**Tip:** Substitution is often a powerful tool for simplifying integrals when the argument of a function is a linear expression.

Learn how to solve the integral ∫sin(3 − t) dt using substitution. This problem introduces key calculus concepts, such as substitution in integration, and provides a step-by-step solution. Suitable for Grades 11-12 and university students.

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