Single Variable Calculus, Chapter 5, 5.5, Section 5.5, Problem 22

Find the indefinite integral $\displaystyle \int \frac{\displaystyle \cos \left( \frac{\pi}{x} \right)}{x^2} dx$

If we let $\displaystyle u = \frac{\pi}{x}$, then $\displaystyle du = \frac{- \pi}{x^2} dx$, so $\displaystyle \frac{1}{x^2} dx = \frac{-1}{\pi} du$. And


$
\begin{equation}
\begin{aligned}

\int \frac{\displaystyle \cos \left( \frac{\pi}{x} \right)}{x^2} dx =& \int \cos \left( \frac{\pi}{x} \right) \frac{1}{x^2} dx
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\int \frac{\displaystyle \cos \left( \frac{\pi}{x} \right)}{x^2} dx =& \int \cos u \frac{-du}{\pi}
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\int \frac{\displaystyle \cos \left( \frac{\pi}{x} \right)}{x^2} dx =& \frac{-1}{\pi} \int \cos u du
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\int \frac{\displaystyle \cos \left( \frac{\pi}{x} \right)}{x^2} dx =& \frac{-1}{\pi} \sin u + C
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\int \frac{\displaystyle \cos \left( \frac{\pi}{x} \right)}{x^2} dx =& \frac{-1}{\pi} \sin \left( \frac{\pi}{x} \right) + C



\end{aligned}
\end{equation}
$