Question

Find the value of \[ \int_0^{\pi/2} x \cos x \, dx. \]

Hint:

Use integration by parts when integrating the product of a polynomial and a trigonometric function.

Answer 1

Use integration by parts: \[ \int u \, dv = uv - \int v \, du, \] where \[ u = x \implies du = dx, \quad dv = \cos x \, dx \implies v = \sin x. \] Then, \[ \int_0^{\pi/2} x \cos x \, dx = \left. x \sin x \right|_0^{\pi/2} - \int_0^{\pi/2} \sin x \, dx. \] Calculate each term: \[ \left. x \sin x \right|_0^{\pi/2} = \frac{\pi}{2} \times 1 - 0 = \frac{\pi}{2}, \] and \[ \int_0^{\pi/2} \sin x \, dx = \left. -\cos x \right|_0^{\pi/2} = (-\cos \frac{\pi}{2}) - (-\cos 0) = 0 + 1 = 1. \] Therefore, \[ \int_0^{\pi/2} x \cos x \, dx = \frac{\pi}{2} - 1. \]
Final answer: \[ \boxed{ \frac{\pi}{2} - 1. } \]