Tìm nguyên hàm của hàm số \(f\left( x \right) = x\sin x\cos x.\)

Tìm nguyên hàm của hàm số \(f\left( x \right) = x\sin x\cos x.\)
A. \(\int {f(x)dx = } \frac{1}{2}\left( {\frac{1}{4}\sin 2x + \frac{x}{2}\cos 2x} \right) + C.\)
B. \(\int {f(x)dx = } - \frac{1}{2}\left( {\frac{1}{4}\sin 2x - \frac{x}{2}\cos 2x} \right) + C.\)
C. \(\int {f(x)dx = } \frac{1}{2}\left( {\frac{1}{4}\sin 2x - \frac{x}{2}\cos 2x} \right) + C.\)
D. \(\int {f(x)dx = } - \frac{1}{2}\left( {\frac{1}{4}\sin 2x + \frac{x}{2}\cos 2x} \right) + C.\)

Ta có: \(f\left( x \right) = \frac{1}{2}x.\sin 2x\)
Ta tính \(I = \frac{1}{2}\int {x.\sin 2x{\rm{d}}x}\).
Đặt: \(\left\{ \begin{array}{l} u = x\\ {\rm{d}}v = \sin 2x \end{array} \right. \Rightarrow \left\{ \begin{array}{l} {\rm{d}}u = {\rm{d}}x\\ v = - \frac{1}{2}\cos 2x \end{array} \right.\)
\(\begin{array}{l} I = \frac{1}{2}\left( { - \frac{1}{2} \cdot x.\cos 2x - \int {\left( { - \frac{1}{2}} \right)\cos 2x{\rm{d}}x} } \right)\\ = \frac{1}{2}\left( { - \frac{1}{2}x.\cos 2x + \frac{1}{2} \cdot \frac{1}{2}\sin 2x} \right) + C \end{array}\)
\(= \frac{1}{2}\left( {\frac{1}{4}\sin 2x - \frac{1}{2}x.\cos 2x} \right) + C.\)