JEE Main 2021 (Online) 22nd July Evening ShiftQuestion 1
If $\int_{0}^{100\pi}\frac{\sin^{2}x}{e^{\left(\frac{x}{\pi}-\left[\frac{x}{\pi}\right]\right)}}dx=\frac{\alpha\pi^{3}}{1+4\pi^{2}}$, $\alpha\in\mathbf{R}$, where $[x]$ is the greatest integer less than or equal to $x$, then the value of $\alpha$ is :
Definite Integration
Solution
Concept. $\sin^2x$ and $\frac x\pi-\big[\frac x\pi\big]$ both have period $\pi$, so the integral is $100$ times the integral over $[0,\pi]$, where $\big[\frac x\pi\big]=0$. Formulas used. (1) $\sin^2x=\frac{1-\cos2x}2$(2) $\int e^{ax}\cos bx\,dx=\dfrac{e^{ax}(a\cos bx+b\sin bx)}{a^2+b^2}$Step 1 - Periodicity. $I=100\int_0^\pi e^{-x/\pi}\sin^2x\,dx$. Step 2 - Two integrals. $\int_0^\pi e^{-x/\pi}dx=\pi(1-e^{-1})$. By (2) with $a=-\frac1\pi$, $b=2$: $$\int_0^\pi e^{-x/\pi}\cos2x\,dx=\frac{\frac1\pi(1-e^{-1})}{\frac1{\pi^2}+4}=\frac{\pi(1-e^{-1})}{1+4\pi^2}.$$ Step 3 - Combine with (1). $$I=50\pi(1-e^{-1})\Big[1-\frac1{1+4\pi^2}\Big]=\frac{200(1-e^{-1})\,\pi^3}{1+4\pi^2}.$$ So $\alpha=200(1-e^{-1})$. Answer: (D)