Đáp án:

• \( I  = \int x \frac{(1 - \cos 2x)}{2} \, dx \)

• Đặt  \( \begin{cases} 
u = \frac{x}{2} \\ 
dv = (1 - \cos 2x) \, dx 
\end{cases} \quad \Rightarrow \quad 
\begin{cases} 
du = \frac{1}{2} \, dx \\ 
v = x - \frac{1}{2} \sin 2x 
\end{cases} \)

• \( I = \frac{x}{2} \left(x - \frac{1}{2} \sin 2x \right) - \frac{1}{2} \int \left(x - \frac{1}{2} \sin 2x \right) \, dx \)

   \(= \frac{x^2}{2} - \frac{x}{4} \sin 2x - \frac{x}{2}+ \frac{1}{4} \sin 2x + C \)