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**Prove It** First of all, your integral has been set up wrong... the volume of a region rotated about the $\displaystyle \displaystyle x$ axis is given by $\displaystyle \displaystyle \pi\int_a^b{[f(x)]^2\,dx}$, so in this case

$\displaystyle \displaystyle \begin{align*} V &= \pi\int_{\frac{2}{3\pi}}^{\frac{2}{\pi}}{\left[\frac{1}{x}\cos{\left(\frac{1}{x}\right)}\right]^2\,dx} \\ &= \pi\int_{\frac{2}{3\pi}}^{\frac{2}{\pi}}{\frac{1}{ x^2}\cos^2{\left(\frac{1}{x}\right)}\,dx} \\ &= -\pi\int_{\frac{2}{3\pi}}^{\frac{2}{\pi}}{-\frac{1}{x^2}\cos^2{\left(\frac{1}{x}\right)}\,dx} \end{align*}$

Now make the substitution $\displaystyle \displaystyle u = \frac{1}{x} \implies du = -\frac{1}{x^2}\,dx$, and note that when $\displaystyle \displaystyle u\left(\frac{2}{3\pi}\right) = \frac{3\pi}{2}$ and $\displaystyle \displaystyle u\left(\frac{2}{\pi}\right) = \frac{\pi}{2}$ so that the integral becomes

$\displaystyle \displaystyle \begin{align*} -\pi\int_{\frac{3\pi}{2}}^{\frac{\pi}{2}}{\cos^2{(u )}\,du} &= \pi\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}{\cos^2{(u )}\,du} \\ &= \pi\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}{\frac{1}{ 2} + \frac{1}{2}\cos{(2u)}\,du} \end{align*}$

Go from here.