Following the hint, we let u = 1 x {\displaystyle u={\frac {1}{x}}} . Then, we have d u = − 1 x 2 d x {\displaystyle du=-{\frac {1}{x^{2}}}dx} and
∫ 1 2 1 x 2 cos ( π x ) d x = ∫ x = 1 x = 2 1 x 2 cos ( π u ) 1 d u / d x d u = ∫ x = 1 x = 2 1 x 2 cos ( π u ) 1 − 1 x 2 d u = ∫ x = 1 x = 2 − cos ( π u ) d u = − 1 π sin ( π u ) | x = 1 x = 2 = − 1 π sin ( π x ) | 1 2 = − 1 π sin ( π 2 ) + 1 π sin ( π ) = − 1 π . {\displaystyle {\begin{aligned}\int _{1}^{2}{\frac {1}{x^{2}}}\cos \left({\frac {\pi }{x}}\right)dx&=\int _{x=1}^{x=2}{\frac {1}{x^{2}}}\cos(\pi u){\frac {1}{du/dx}}\,du\\&=\int _{x=1}^{x=2}{\frac {1}{x^{2}}}\cos(\pi u){\frac {1}{-{\frac {1}{x^{2}}}}}\,du\\&=\int _{x=1}^{x=2}-\cos(\pi u)du\\&=\left.-{\frac {1}{\pi }}\sin(\pi u)\right|_{x=1}^{x=2}\\&=\left.-{\frac {1}{\pi }}\sin \left({\frac {\pi }{x}}\right)\right|_{1}^{2}\\&=-{\frac {1}{\pi }}\sin \left({\frac {\pi }{2}}\right)+{\frac {1}{\pi }}\sin(\pi )\\&=-{\frac {1}{\pi }}.\end{aligned}}}
Answer: − 1 π {\displaystyle \color {blue}-{\frac {1}{\pi }}}
. Then, we have 
and
