Ta có: \(\int x\sin ^{2} x\, {\rm d}x =\frac{1}{2} \int x\left(1-\cos 2x\right)\, {\rm d}x =\frac{1}{2} \int x\, {\rm d}x-\frac{1}{2} \int x\cos 2x\, {\rm d}x\)
Đặt: \(\left\{\begin{array}{l} {u=x} \\ {{\rm d}v={\rm co}s2x{\rm d}x} \end{array}\right. \Rightarrow \left\{\begin{array}{l} {{\rm d}u={\rm d}x} \\ {v=\frac{1}{2} \sin 2x} \end{array}\right.\)
\(\Rightarrow \int x.\cos 2x\, {\rm d}x =\frac{1}{2} x\sin 2x-\frac{1}{2} \int \sin 2x\, {\rm d}x= \frac{1}{2} x\sin 2x+\frac{1}{4} \cos 2x+C_{1}\)
Vậy \(\int x.\sin ^{2} x\, {\rm d}x =\frac{1}{4} x^{2} -\frac{1}{4} x\sin 2x-\frac{1}{8} {\rm cos}2x+C\)