# Solve this

Question:

If $\int_{0}^{100 \pi} \frac{\sin ^{2} x}{e^{\left(\frac{x}{\pi}-\left[\frac{x}{\pi}\right]\right)}} d x=\frac{\alpha \pi^{3}}{1+4 \pi^{2}}, \alpha \in \mathbf{R}$ where $[\mathrm{x}]$ is the greatest integer less than or equal to $x$, then the value of $\alpha$ is :

1. $200\left(1-\mathrm{e}^{-1}\right)$

2. $100(1-\mathrm{e})$

3. $50(\mathrm{e}-1)$

4. $150\left(\mathrm{e}^{-1}-1\right)$

Correct Option: 1

Solution:

$I=\int_{0}^{100 \pi} \frac{\sin ^{2} x}{e^{\{x / \pi\}}} d x=100 \int_{0}^{\pi} \frac{\sin ^{2} x}{e^{x / \pi}} d x$

$100 \int_{0}^{\pi} e^{-x / \pi} \frac{(1-\cos 2 x)}{2} d x$

$=50\left\{\int_{0}^{\pi} \mathrm{e}^{-\mathrm{x} / \pi} \mathrm{d} \mathrm{x}-\int_{0}^{\pi} \mathrm{e}^{-\mathrm{x} / \pi} \cos 2 \mathrm{x} \mathrm{dx}\right\}$

$I_{1}=\int_{0}^{\pi} e^{-x / \pi} d x=\left[-\pi e^{-x / \pi}\right]_{0}^{\pi}=\pi\left(1-e^{-1}\right)$

$I_{2}=\int_{0}^{\pi} e^{-x / \pi} \cos 2 x d x$

$\left.=-\pi \mathrm{e}^{-x / \pi} \cos 2 x\right]_{0}^{\pi}-\int-\pi \mathrm{e}^{-x / \pi}(-2 \sin 2 x) d x$

$=\pi\left(1-\mathrm{e}^{-1}\right)-2 \pi \int_{0}^{\pi} \mathrm{e}^{-x / \pi} \sin 2 x d x$

$\left.=\pi\left(1-\mathrm{e}^{-1}\right)-2 \pi\left\{-\pi \mathrm{e}^{-\mathrm{x} / \pi} \sin 2 \mathrm{x}\right]_{0}^{\pi}-\int_{0}^{\pi}-\pi \mathrm{e}^{-\mathrm{x} / \pi} 2 \cos 2 \mathrm{xd} \mathrm{x}\right\}$

$=\pi\left(1-e^{-1}\right)-4 \pi^{2} I_{2}$

$\Rightarrow \mathrm{I}_{2}=\frac{\pi\left(1-\mathrm{e}^{-1}\right)}{1+4 \pi^{2}}$

$\therefore \mathrm{I}=50\left\{\pi\left(1-\mathrm{e}^{-1}\right)-\frac{\pi\left(1-\mathrm{e}^{-1}\right)}{1+4 \pi^{2}}\right\}$

$=\frac{200\left(1-\mathrm{e}^{-1}\right) \pi^{3}}{1+4 \pi^{2}}$