$\int_{0}^{19 / 2} \frac{x^{2}}{\sqrt{361-x^{2}}} d x$

Step 1 ：First, we recognize that the integral is in the form of \(\int \frac{x^{2}}{\sqrt{a^{2}-x^{2}}} dx\), which is a standard integral form.

Step 2 ：We know that the standard integral form can be solved by substituting \(x = a \sin \theta\).

Step 3 ：Let's substitute \(x = 19 \sin \theta\) into the integral. Then, \(dx = 19 \cos \theta d\theta\).

Step 4 ：Substitute these values into the integral, we get \(\int_{0}^{\pi / 2} \frac{(19 \sin \theta)^{2}}{\sqrt{361-(19 \sin \theta)^{2}}} \cdot 19 \cos \theta d\theta\).

Step 5 ：Simplify the integral to \(\int_{0}^{\pi / 2} 19^{3} \sin^{2} \theta \cos \theta d\theta\).

Step 6 ：Next, we use the double-angle formula \(\sin^{2} \theta = \frac{1 - \cos 2\theta}{2}\) to simplify the integral.

Step 7 ：Substitute \(\sin^{2} \theta\) with \(\frac{1 - \cos 2\theta}{2}\) in the integral, we get \(\int_{0}^{\pi / 2} \frac{19^{3}}{2} (1 - \cos 2\theta) \cos \theta d\theta\).

Step 8 ：Split the integral into two parts, we get \(\frac{19^{3}}{2} \int_{0}^{\pi / 2} \cos \theta d\theta - \frac{19^{3}}{2} \int_{0}^{\pi / 2} \cos 2\theta \cos \theta d\theta\).

Step 9 ：The first integral can be solved directly, and the second integral can be solved by using the product-to-sum formula \(\cos A \cos B = \frac{1}{2} [\cos(A - B) + \cos(A + B)]\).

Step 10 ：Substitute \(\cos A \cos B\) with \(\frac{1}{2} [\cos(A - B) + \cos(A + B)]\) in the second integral, we get \(-\frac{19^{3}}{4} \int_{0}^{\pi / 2} [\cos \theta + \cos 3\theta] d\theta\).

Step 11 ：Calculate the two integrals separately, we get \(\frac{19^{3}}{2} [\sin \theta]_{0}^{\pi / 2} - \frac{19^{3}}{4} [\sin \theta + \frac{1}{3} \sin 3\theta]_{0}^{\pi / 2}\).

Step 12 ：Substitute the upper and lower limits into the integral, we get \(\frac{19^{3}}{2} \cdot 1 - \frac{19^{3}}{4} \cdot (1 + \frac{1}{3})\).

Step 13 ：Simplify the expression, we get \(\frac{19^{3}}{2} - \frac{19^{3}}{4} - \frac{19^{3}}{12}\).

Step 14 ：Combine like terms, we get \(\frac{19^{3}}{4}\).

Step 15 ：So, the final answer is \(\boxed{\frac{19^{3}}{4}}\).