Answer
$-\ln |\sin \dfrac{1}{x}|+C$
Work Step by Step
Our aim is to solve the integral $ \int \dfrac{\cos (1/x)}{x^2 \sin (1/x)}\ dx=\int \cot (\dfrac{1}{x}) (\dfrac{1}{x^2}) \ dx $
Let us consider that $a =\dfrac{1}{x}$ and $da=\dfrac{-1}{x^2} \ dx$
Now, $ \int \cot (\dfrac{1}{x}) (\dfrac{1}{x^2}) \ dx = \int \cot a (-da)$
or, $= -\ln |\sin a|+C $
or, $=-\ln |\sin \dfrac{1}{x}|+C$
