We need to evaluate the definite integral $J = \int_{0}^{1} x^4 (1 - x^2)^{3/2} dx$.

1. Problem Description

We need to evaluate the definite integral $J = \int_{0}^{1} x^4 (1 - x^2)^{3/2} dx$.

2. Solution Steps

We will use the trigonometric substitution $x = \sin(\theta)$. Then $dx = \cos(\theta) d\theta$.

When $x=0$, $\theta = \arcsin(0) = 0$. When $x=1$, $\theta = \arcsin(1) = \pi/2$.

The integral becomes:

$J = \int_{0}^{\pi/2} \sin^4(\theta) (1 - \sin^2(\theta))^{3/2} \cos(\theta) d\theta = \int_{0}^{\pi/2} \sin^4(\theta) (\cos^2(\theta))^{3/2} \cos(\theta) d\theta = \int_{0}^{\pi/2} \sin^4(\theta) \cos^3(\theta) \cos(\theta) d\theta = \int_{0}^{\pi/2} \sin^4(\theta) \cos^4(\theta) d\theta$.

We can rewrite the integral as:

$J = \int_{0}^{\pi/2} (\sin(\theta) \cos(\theta))^4 d\theta = \int_{0}^{\pi/2} (\frac{1}{2} \sin(2\theta))^4 d\theta = \frac{1}{16} \int_{0}^{\pi/2} \sin^4(2\theta) d\theta$.

Let $u = 2\theta$, then $du = 2 d\theta$, so $d\theta = \frac{1}{2} du$.

When $\theta = 0$, $u = 0$. When $\theta = \pi/2$, $u = \pi$.

The integral becomes:

$J = \frac{1}{16} \int_{0}^{\pi} \sin^4(u) \frac{1}{2} du = \frac{1}{32} \int_{0}^{\pi} \sin^4(u) du$.

Since $\sin^4(u)$ is symmetric about $u = \pi/2$, we have

$\int_{0}^{\pi} \sin^4(u) du = 2 \int_{0}^{\pi/2} \sin^4(u) du$.

So, $J = \frac{1}{32} \cdot 2 \int_{0}^{\pi/2} \sin^4(u) du = \frac{1}{16} \int_{0}^{\pi/2} \sin^4(u) du$.

We use the reduction formula for $\int_{0}^{\pi/2} \sin^n(x) dx$:

$\int_{0}^{\pi/2} \sin^n(x) dx = \frac{n-1}{n} \int_{0}^{\pi/2} \sin^{n-2}(x) dx$.

Thus, $\int_{0}^{\pi/2} \sin^4(u) du = \frac{4-1}{4} \int_{0}^{\pi/2} \sin^2(u) du = \frac{3}{4} \int_{0}^{\pi/2} \sin^2(u) du$.

Now, $\int_{0}^{\pi/2} \sin^2(u) du = \frac{2-1}{2} \int_{0}^{\pi/2} \sin^0(u) du = \frac{1}{2} \int_{0}^{\pi/2} 1 du = \frac{1}{2} [\theta]_0^{\pi/2} = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}$.

So, $\int_{0}^{\pi/2} \sin^4(u) du = \frac{3}{4} \cdot \frac{\pi}{4} = \frac{3\pi}{16}$.

Therefore, $J = \frac{1}{16} \cdot \frac{3\pi}{16} = \frac{3\pi}{256}$.

3. Final Answer

$\frac{3\pi}{256}$