valuate 
9
sin(x) cos(x) dx

 by u sub, give me four answers in order. first u = cos(x), then u = sin(x),. After those do it by identifying sin(2x) = 2sin(x)cos(x), finally do it by integrate by parts

Let's evaluate the integral \( I = \int 9 \sin(x) \cos(x) \, dx \) using four different methods, as requested:

### Method 1: Substitution with \( u = \cos(x) \)

1. **Substitute** \( u = \cos(x) \), hence \( du = -\sin(x) dx \).

   The integral becomes:
   \[
   I = \int 9 \sin(x) \cos(x) \, dx = -9 \int u \, du
   \]

2. **Integrate**:
   \[
   -9 \int u \, du = -9 \left( \frac{u^2}{2} \right) + C = -\frac{9}{2} \cos^2(x) + C
   \]

   **Answer 1**: \( I = -\frac{9}{2} \cos^2(x) + C \)

---

### Method 2: Substitution with \( u = \sin(x) \)

1. **Substitute** \( u = \sin(x) \), hence \( du = \cos(x) dx \).

   The integral becomes:
   \[
   I = \int 9 u \, du
   \]

2. **Integrate**:
   \[
   9 \int u \, du = 9 \left( \frac{u^2}{2} \right) + C = \frac{9}{2} \sin^2(x) + C
   \]

   **Answer 2**: \( I = \frac{9}{2} \sin^2(x) + C \)

---

### Method 3: Using the identity \( \sin(2x) = 2 \sin(x) \cos(x) \)

We recognize that \( \sin(2x) = 2 \sin(x) \cos(x) \), so:
\[
9 \sin(x) \cos(x) = \frac{9}{2} \sin(2x)
\]

Now, integrate:
\[
I = \int \frac{9}{2} \sin(2x) \, dx
\]

The integral of \( \sin(2x) \) is \( -\frac{1}{2} \cos(2x) \), so:
\[
I = \frac{9}{2} \left( -\frac{1}{2} \cos(2x) \right) + C = -\frac{9}{4} \cos(2x) + C
\]

**Answer 3**: \( I = -\frac{9}{4} \cos(2x) + C \)

---

### Method 4: Integration by Parts

We can apply integration by parts where we let \( u = \sin(x) \) and \( dv = 9 \cos(x) \, dx \). Then, \( du = \cos(x) \, dx \) and \( v = 9 \sin(x) \).

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \), we get:

1. \( u = \sin(x) \), \( dv = 9 \cos(x) \, dx \)
2. \( v = 9 \sin(x) \), \( du = \cos(x) \, dx \)

Now:
\[
I = 9 \sin(x) \sin(x) - \int 9 \sin(x) \cos(x) \, dx
\]

This brings us back to the original integral on the right-hand side. So we use algebra to solve:
\[
I = \frac{9}{2} \sin^2(x) + C
\]

**Answer 4**: \( I = \frac{9}{2} \sin^2(x) + C \)

---

### Summary of Answers:
1. \( I = -\frac{9}{2} \cos^2(x) + C \)
2. \( I = \frac{9}{2} \sin^2(x) + C \)
3. \( I = -\frac{9}{4} \cos(2x) + C \)
4. \( I = \frac{9}{2} \sin^2(x) + C \)

Would you like further details or clarification on any of these methods?

---

**5 Related Questions**:
1. How do you choose an appropriate substitution for integration?
2. What is the integration by parts formula, and how is it applied?
3. Can we generalize the identity \( \sin(2x) = 2 \sin(x) \cos(x) \) for higher powers of sine and cosine?
4. How does the symmetry of sine and cosine functions affect integration?
5. What is the geometric interpretation of \( \sin(2x) \) in integrals?

**Tip**: Substituting trigonometric identities can often simplify complex integrals and make them easier to compute!

Evaluate 9 sin(x) cos(x) dx by substitution, providing four answers in order: first u = cos(x), then u = sin(x), then using sin(2x) = 2sin(x)cos(x), and finally by integration by parts.

Explore how to evaluate the integral of 9 sin(x) cos(x) dx using multiple methods, including substitution (u = cos(x) and u = sin(x)), the identity sin(2x) = 2 sin(x) cos(x), and integration by parts. A detailed step-by-step solution suitable for advanced high school and college calculus students.

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