= Solution
The squared <norm> of the original function is
$$
\int_{\mathbb R}|f(x)|^2\,dx=\int_{-\pi/2}^{\pi/2}\sin^2x\,dx=\frac\pi2.
$$
Applying the <Parseval identity> to part (c), using the evenness of the squared transform, gives
$$
\frac\pi2=\frac1{2\pi}\int_{\mathbb R}\frac{4k^2\cos^2(\pi k/2)}{(k^2-1)^2}\,dk=\frac4\pi\int_0^{\infty}\frac{k^2\cos^2(\pi k/2)}{(k^2-1)^2}\,dk.
$$
Thus the last <integral> equals $\pi^2/8$. Now set $u=\pi k/2$. Since
$$
\frac{k^2\,dk}{(k^2-1)^2}=\frac\pi2\frac{u^2\,du}{(u^2-\pi^2/4)^2},
$$
we obtain
$$
\boxed{\int_0^{\infty}\frac{u^2\cos^2u}{(u^2-\pi^2/4)^2}\,du=\frac2\pi\frac{\pi^2}{8}=\frac\pi4.}
$$
The apparent singularity at $u=\pi/2$ is removable: the continuous value of the integrand there is $1/4$. No principal-value interpretation is needed.
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