The given integral is:

$2 \int_0^1 x^3 \sqrt{1 - x^2} \, dx$

This is a standard integral that can be solved using a substitution method. Let's walk through the steps.

Step 1: Substitution

Let’s use the substitution $x = \sin(\theta)$. Therefore, $dx = \cos(\theta) \, d\theta$, and since $1 - x^2 = \cos^2(\theta)$, we have:

$\sqrt{1 - x^2} = \cos(\theta)$

Now, changing the limits:

• When $x = 0$, $\sin(\theta) = 0$, so $\theta = 0$.
• When $x = 1$, $\sin(\theta) = 1$, so $\theta = \frac{\pi}{2}$.

Thus, the integral becomes:

$2 \int_0^{\frac{\pi}{2}} (\sin^3(\theta)) (\cos(\theta)) (\cos(\theta)) \, d\theta$

This simplifies to:

$2 \int_0^{\frac{\pi}{2}} \sin^3(\theta) \cos^2(\theta) \, d\theta$

Step 2: Simplify the Expression

We now need to simplify $\sin^3(\theta) \cos^2(\theta)$. Using the identity $\cos^2(\theta) = 1 - \sin^2(\theta)$, we get:

$\sin^3(\theta) \cos^2(\theta) = \sin^3(\theta) (1 - \sin^2(\theta))$

Thus, the integral becomes:

$2 \int_0^{\frac{\pi}{2}} \sin^3(\theta) (1 - \sin^2(\theta)) \, d\theta$

Expanding the terms:

$2 \int_0^{\frac{\pi}{2}} \left( \sin^3(\theta) - \sin^5(\theta) \right) \, d\theta$

Step 3: Separate the Terms

Now, we can break the integral into two separate integrals:

$2 \left( \int_0^{\frac{\pi}{2}} \sin^3(\theta) \, d\theta - \int_0^{\frac{\pi}{2}} \sin^5(\theta) \, d\theta \right)$

Step 4: Use Standard Integrals

The integrals of $\sin^n(\theta)$ for odd powers can be computed using known formulas. The general formula for $\int_0^{\frac{\pi}{2}} \sin^{2n+1}(\theta) \, d\theta$ is:

$\int_0^{\frac{\pi}{2}} \sin^{2n+1}(\theta) \, d\theta = \frac{4^n}{(2n+1) \cdot 2^{2n}} \cdot \frac{\pi}{2}$

For $\sin^3(\theta)$:

$\int_0^{\frac{\pi}{2}} \sin^3(\theta) \, d\theta = \frac{4}{3}$

For $\sin^5(\theta)$:

$\int_0^{\frac{\pi}{2}} \sin^5(\theta) \, d\theta = \frac{8}{5}$

Step 5: Final Calculation

Substitute these values back into the equation:

$2 \left( \frac{4}{3} - \frac{8}{5} \right) = 2 \left( \frac{20}{15} - \frac{24}{15} \right) = 2 \times \frac{-4}{15} = \frac{-8}{15}$

Final Answer:

$\boxed{-\frac{8}{15}}$

Would you like any further explanation or breakdown of the steps? Here are 5 related questions to explore:

1. How do you handle integrals with powers of trigonometric functions?
2. Can you explain how the substitution $x = \sin(\theta)$ simplifies the integral?
3. What is the general method for solving integrals of the form $\int_0^{\frac{\pi}{2}} \sin^n(\theta) \, d\theta$?
4. How do you apply standard integral tables to evaluate trigonometric integrals?
5. What are some other types of integrals that benefit from trigonometric substitution?

Tip: For integrals involving trigonometric functions, often substitution and identity manipulations can reduce the problem to simpler forms, making the solution more straightforward.