\(\int \left(2x-1\right)\cos ^{2} x\, {\rm d}x .\)
Đặt \(\left\{\begin{array}{l} {u=2x-1} \\ {{\rm d}v=\cos ^{2} x{\rm d}x=\frac{1}{2} \left(1+\cos 2x\right){\rm d}x} \end{array}\right. \Rightarrow \left\{\begin{array}{l} {{\rm d}u=2{\rm d}x} \\ {v=\frac{1}{2} \left(x+\frac{1}{2} \sin 2x\right)} \end{array}\right. \)
Khi đó \(\int \left(2x-1\right)\cos ^{2} x{\rm d}x =\frac{1}{4} \left(2x-1\right)\left(2x+\sin 2x\right)-\int \left(x+\frac{1}{2} \sin 2x\right){\rm d}x
=\frac{1}{4} \left(2x-1\right)\left(2x+\sin 2x\right)-\frac{x^{2} }{2} +\frac{1}{4} \cos 2x+C.\)