To integrate \( \int 3x^3 \sqrt{16 - x^2} \, dx \), we can use trigonometric substitution.

### Step 1: Substitution
To handle the square root term \( \sqrt{16 - x^2} \), we use the substitution:

\[
x = 4 \sin \theta \quad \Rightarrow \quad dx = 4 \cos \theta \, d\theta
\]

This substitution simplifies \( \sqrt{16 - x^2} \) because:

\[
\sqrt{16 - x^2} = \sqrt{16 - 16 \sin^2 \theta} = 4 \cos \theta
\]

### Step 2: Substitute into the integral
Substitute \( x = 4 \sin \theta \) and \( dx = 4 \cos \theta \, d\theta \) into the integral:

\[
\int 3x^3 \sqrt{16 - x^2} \, dx = \int 3(4 \sin \theta)^3 (4 \cos \theta)(4 \cos \theta \, d\theta)
\]

Simplify the expression:

\[
= 3 \times 64 \int \sin^3 \theta \cos^2 \theta \, d\theta
\]

This simplifies to:

\[
192 \int \sin^3 \theta \cos^2 \theta \, d\theta
\]

### Step 3: Use trigonometric identities
We can use the identity \( \sin^2 \theta = 1 - \cos^2 \theta \) to simplify \( \sin^3 \theta \). First, rewrite \( \sin^3 \theta = \sin \theta (1 - \cos^2 \theta) \):

\[
192 \int \sin \theta (1 - \cos^2 \theta) \cos^2 \theta \, d\theta
\]

Now distribute:

\[
192 \int \sin \theta (\cos^2 \theta - \cos^4 \theta) \, d\theta
\]

This separates into two integrals:

\[
192 \left( \int \sin \theta \cos^2 \theta \, d\theta - \int \sin \theta \cos^4 \theta \, d\theta \right)
\]

### Step 4: Solve each integral
1. For \( \int \sin \theta \cos^2 \theta \, d\theta \), we can use substitution:

   Let \( u = \cos \theta \), so \( du = -\sin \theta \, d\theta \). The integral becomes:

   \[
   -\int u^2 \, du = -\frac{u^3}{3} = -\frac{\cos^3 \theta}{3}
   \]

2. For \( \int \sin \theta \cos^4 \theta \, d\theta \), use the same substitution \( u = \cos \theta \):

   \[
   -\int u^4 \, du = -\frac{u^5}{5} = -\frac{\cos^5 \theta}{5}
   \]

### Step 5: Combine the results
Substitute these back into the integral:

\[
192 \left( -\frac{\cos^3 \theta}{3} + \frac{\cos^5 \theta}{5} \right)
\]

Factor out \( -192 \):

\[
-192 \left( \frac{\cos^3 \theta}{3} - \frac{\cos^5 \theta}{5} \right)
\]

### Step 6: Back-substitute \( \theta \)
Since \( x = 4 \sin \theta \), we have \( \cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \frac{x^2}{16}} \). Thus:

\[
\cos \theta = \frac{\sqrt{16 - x^2}}{4}
\]

Substitute this back into the expression:

\[
-192 \left( \frac{\left( \frac{\sqrt{16 - x^2}}{4} \right)^3}{3} - \frac{\left( \frac{\sqrt{16 - x^2}}{4} \right)^5}{5} \right)
\]

Simplify the powers of \( \frac{\sqrt{16 - x^2}}{4} \) and combine terms to get the final answer.

Would you like more detailed steps on the simplification, or to discuss another topic related to this? Here are 5 questions that can deepen understanding:

1. How does trigonometric substitution work for integrals with square roots?
2. What are common substitutions for integrals involving square roots of quadratic terms?
3. Why do we factor powers of sine or cosine in certain integrals?
4. How can we simplify higher powers of trigonometric functions?
5. What is the process for back-substituting in trigonometric integrals?

**Tip:** When faced with complex trigonometric integrals, simplifying using identities like \( \sin^2 \theta + \cos^2 \theta = 1 \) often makes the process smoother.

Learn how to integrate the function 3x^3 sqrt(16 - x^2) using trigonometric substitution. This solution involves substituting x = 4 sin(θ) and using trigonometric identities to simplify the integral. Suitable for undergraduate calculus students.

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