Question

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

• A. $50(e-1)$
• B. $100(1-e)$
• C. $150(e^{-1}-1)$
• D. $200(1-e^{-1})$

Answer 1

The correct option is D $200(1-e^{-1})$

$I=\underset{0}{\overset{100\pi }{\int }}\frac{{\sin }^{2}x}{e^{\frac{x}{\pi }-\left[\frac{x}{\pi }\right]}}dx$
$\because$ Integrand is periodic with period $1$
$\therefore I=100\underset{0}{\overset{\pi }{\int }}\frac{{\sin }^{2}x}{e^{\left\{\frac{x}{\pi }\right\}}}dx$
Let $\frac{x}{\pi }=t\Rightarrow dx=\pi dt$
$\Rightarrow I=100\pi \underset{0}{\overset{1}{\int }}\frac{{\sin }^2\left(\pi t\right)dt}{e^t}$
$=50\pi \underset{0}{\overset{1}{\int }}{e}^{-t}(1-\cos 2\pi t)dt$
$=50\pi \underset{0}{\overset{1}{\int }}{e}^{-t}dt-50\pi \underset{0}{\overset{1}{\int }}{e}^{-t}\cos \left(2\pi t\right)dt$
$=-50\pi {\left[e^{-t}\right]}_0^1$
$-50\pi {\left[\frac{e^{-t}}{1+4{\pi }^2}\right(-\cos 2\pi t+2\pi \sin 2\pi t\left)\right]}_0^1$
$(\because \int e^{ax}\cdot \cos bxdx=\frac{e^{ax}}{a^2+b^2}(a\cos bx+b\sin bx)+c)$
$=-50\pi (e^{-1}-1)-\frac{50\pi }{1+4{\pi }^2}\left(e^{-1}\right(-1+0)-(-1+0\left)\right)$
$=-50\pi (e^{-1}-1)-\frac{50\pi }{1+4{\pi }^2}(1-e^{-1})$
$=-50\pi (e^{-1}-1)-\frac{50\pi (1-e^{-1})}{1+4{\pi }^2}$
$=\frac{200{\pi }^3(1-e^{-1})}{1+4{\pi }^2}=\frac{\alpha {\pi }^3}{1+4{\pi }^3}$ (Given)
$\therefore \alpha =200(1-e^{-1})$