Đặt \(\left\{\begin{array}{l} {u=\cos \left(\ln x\right)} \\ {{\rm d}v={\rm d}x} \end{array}\right. \Rightarrow \left\{\begin{array}{l} {{\rm d}u=-\frac{1}{x} \sin \left(\ln x\right){\rm d}x} \\ {v=x} \end{array}\right. \)
\(\Rightarrow \int \cos \left(\ln x\right){\rm d}x =x\cos \left(\ln x\right)+\int \sin \left(\ln x\right){\rm d}x .\)
Với \(\int \sin \left(\ln x\right){\rm d}x \)
Đặt \(\left\{\begin{array}{l} {u'=\sin \left(\ln x\right)} \\ {{\rm d}v'={\rm d}x} \end{array}\right. \Rightarrow \left\{\begin{array}{l} {{\rm d}u'=\frac{1}{x} \cos \left(\ln x\right){\rm d}x} \\ {v'=x} \end{array}\right. \)\(
\Rightarrow \int \sin \left(\ln x\right){\rm d}x =x\sin \left(\ln x\right)-\int \cos \left(\ln x\right) {\rm d}x.\)
Vậy \(\int \cos \left(\ln x\right) {\rm d}x=\frac{1}{2} x\left[\cos \left(\ln x\right)+\sin \left(\ln x\right)\right]+C.\)