Dear student I=∫1x2cos21xdxTake u=1x⇒dudx=-1x2So,I=-∫cos2u du=-∫1+cos2u2du=-∫12du-∫12cos2u du=-u2-sin2u4+C=-121x+12sin2x+C

Solve this:

Dear student

I = \int \frac{1}{x^{2}} \cos^{2} (\frac{1}{x}) dx Take u = \frac{1}{x} \Rightarrow \frac{du}{dx} = - \frac{1}{x^{2}} So, I = - \int \cos^{2} u du = - \int \frac{1 + \cos 2 u}{2} du = - \int \frac{1}{2} du - \int \frac{1}{2} \cos 2 u du = - \frac{u}{2} - \frac{\sin 2 u}{4} + C = - \frac{1}{2} (\frac{1}{x} + \frac{1}{2} \sin (\frac{2}{x})) + C