$\int_{0}^{3 \pi / 2} x \sin x d x$

Step 1 ：Given the integral problem \(\int_{0}^{3 \pi / 2} x \sin x d x\), we can solve this using the method of integration by parts.

Step 2 ：The formula for integration by parts is \(\int udv = uv - \int vdu\).

Step 3 ：Let's set \(u = x\) and \(dv = \sin(x) dx\).

Step 4 ：We then find \(du\) and \(v\). \(du\) is the derivative of \(x\) which is \(dx\), and \(v\) is the integral of \(\sin(x)\) which is \(-\cos(x)\).

Step 5 ：Substituting these into the formula, we get \(-x\cos(x) + \int \cos(x) dx\).

Step 6 ：Evaluating the integral, we get \(-x\cos(x) + \sin(x)\).

Step 7 ：Evaluating this from \(0\) to \(3 \pi / 2\), we get \(-1\).

Step 8 ：So, the definite integral of \(x \sin x\) from \(0\) to \(3 \pi / 2\) is \(\boxed{-1}\).