Official solution
\[I = \int_0^{100\pi } {{{{{\sin }^2}x} \over {{e^{\left( {{x \over \pi } - \left[ {{x \over \pi }} \right]} \right)}}}}dx} \]
\[\because\] Integrand is periodic with period 1
\[\therefore\] \[I = 100\int_0^\pi {{{{{\sin }^2}x} \over {{e^{\left\{ {{x \over \pi }} \right\}}}}}} dx\]
Let \[{x \over \pi } = t \Rightarrow dx = \pi dt\]
\[ = 100\pi \int_0^1 {{{{{\sin }^2}(\pi t)dt} \over {{e^t}}}} \]
\[ = 50\pi \int_0^1 {{e^{ - t}}(1 - \cos 2\pi t)dt} \]
\[ = 50\pi \int_0^1 {{e^{ - t}}dt - 50\pi \int_0^1 {{e^{ - t}}} \cos (2\pi t)dt} \]
\[ = - 50\left[ {{e^{ - t}}} \right]_0^1 - 50\pi \left[ {{{{e^{ - t}}} \over {1 + 4{\pi ^2}}}( - \cos 2\pi t + 2\pi \sin 2\pi t)} \right]_0^1\]
\[ = - 50\pi ({e^{ - 1}} - 1) - {{50\pi } \over {1 + 4{\pi ^2}}}({e^{ - 1}}( - 1 + 0) - ( - 1 + 0))\]
\[ = - 50\pi ({e^{ - 1}} - 1) - {{50\pi (1 - {e^{ - 1}})} \over {1 + 4{\pi ^2}}}\]
\[ = {{200{\pi ^3}(1 - {e^{ - 1}})} \over {1 + 4{\pi ^2}}} = {{\alpha {\pi ^3}} \over {1 + 4{\pi ^3}}}\] (Given)
\[\therefore\] \[\alpha = 200(1 - {e^{ - 1}})\]