Correct option is (4) \(200\left(1-e^{-1}\right)\)
\(I=\int\limits_0^{100 \pi} \frac{\sin ^2 x}{e^{\frac{x}{\pi}}\left[\frac{x}{\pi}\right]} d x \)
\(\because \text { Integrand is periodic with period } 1\)
\(\therefore I=100 \int\limits_0^\pi \frac{\sin ^2 x}{\left\{\frac{x}{\pi}\right\}} d x\)
\( \text { Let } \frac{x}{\pi}=t \Rightarrow d x=\pi d t \)
\(\Rightarrow I=100 \pi \int\limits_0^1 \frac{\sin ^2(\pi t) d t}{e^t}\)
\(=50 \pi \int\limits_0^1 e^{-t}(1-\cos 2 \pi t) d t\)
\(=50 \pi \int\limits_0^1 e^{-t} d t-50 \pi \int\limits_0^1 e^{-t} \cos (2 \pi t) d t\)
\(=-50 \pi\left[e^{-t}\right]_0^1\)
\(-50 \pi\left[\frac{e^{-t}}{1+4 \pi^2}(-\cos 2 \pi t+2 \pi \sin 2 \pi t)\right] ^1_0\)
\(\left(\because \int e^{a x} \cdot \cos b x d x=\frac{e^{a x}}{a^2+b^2}(a \cos b x+b \sin b x)+c\right)\)
\(=-50 \pi\left(e^{-1}-1\right)-\frac{50 \pi}{1+4 \pi^2}\left(e^{-1}(-1+0)-(-1+0)\right) \)
\(=-50 \pi\left(e^{-1}-1\right)-\frac{50 \pi}{1+4 \pi^2}\left(1-e^{-1}\right)\)
\(=-50 \pi\left(e^{-1}-1\right)-\frac{50 \pi\left(1-e^{-1}\right)}{1+4 \pi^2} \)
\(=\frac{200 \pi^3\left(1-e^{-1}\right)}{1+4 \pi^2}=\frac{a \pi^3}{1+4 \pi^3}\quad\text{(Given) }\)
\(\therefore a=200\left(1-e^{-1}\right)\)
