We need to evaluate the integral:

$\int \frac{x^3}{\sqrt{16 - x^2}} \, dx$

This is a trigonometric substitution problem. Let's break it down step by step.

Step 1: Substitution

We recognize that the expression under the square root resembles $\sqrt{a^2 - x^2}$, where $a = 4$. In such cases, we use the substitution:

$x = 4 \sin(\theta) \quad \text{which gives} \quad dx = 4 \cos(\theta) \, d\theta$

Substituting into the integral:

• $x^2 = 16 \sin^2(\theta)$
• $\sqrt{16 - x^2} = \sqrt{16(1 - \sin^2(\theta))} = 4 \cos(\theta)$
• $x^3 = (4 \sin(\theta))^3 = 64 \sin^3(\theta)$

Thus, the integral becomes:

$\int \frac{64 \sin^3(\theta)}{4 \cos(\theta)} \cdot 4 \cos(\theta) \, d\theta$

Step 2: Simplify the Expression

The $\cos(\theta)$ terms cancel out, leaving:

$\int 64 \sin^3(\theta) \, d\theta$

Step 3: Use Trigonometric Identity

We can use the identity $\sin^3(\theta) = \sin(\theta) (1 - \cos^2(\theta))$ to simplify further:

$64 \int \sin(\theta) (1 - \cos^2(\theta)) \, d\theta$

Now, break it into two integrals:

$64 \int \sin(\theta) \, d\theta - 64 \int \sin(\theta) \cos^2(\theta) \, d\theta$

Step 4: Solve Each Integral

The first integral is straightforward:

$64 \int \sin(\theta) \, d\theta = -64 \cos(\theta)$

For the second integral, we can use the substitution $u = \cos(\theta), \, du = -\sin(\theta) \, d\theta$:

$64 \int \sin(\theta) \cos^2(\theta) \, d\theta = -64 \int u^2 \, du = -64 \left( \frac{u^3}{3} \right) = -\frac{64}{3} \cos^3(\theta)$

Step 5: Combine the Results

Combining the two integrals, we get:

$-64 \cos(\theta) + \frac{64}{3} \cos^3(\theta)$

Step 6: Substitute Back

Recall that $\cos(\theta) = \sqrt{1 - \frac{x^2}{16}}$ and $\theta = \sin^{-1} \left( \frac{x}{4} \right)$. Therefore, the final result is:

$-64 \sqrt{1 - \frac{x^2}{16}} + \frac{64}{3} \left( \sqrt{1 - \frac{x^2}{16}} \right)^3 + C$

This is the solution to the integral.

Would you like more details or have any questions?

5 Related Questions:

1. How do we decide when to use trigonometric substitution in integrals?
2. Can you explain how to derive the trigonometric substitution formulas?
3. How does substitution simplify integrals involving square roots?
4. What are some common identities used in trigonometric integration?
5. How do we handle definite integrals using trigonometric substitution?

Tip: Whenever you see $\sqrt{a^2 - x^2}$ in an integral, consider using $x = a \sin(\theta)$ as a substitution.