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

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

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

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Definite Integrals

\int_{0}^{100 π} e^{- (\frac{x}{π} - [\frac{x}{π}])} sin^{2} x d x = \frac{α π ^{3}}{1 + 4 π ^{2}}, α \in R

, where

[x]

is the greatest integer less than or equal to

x

, then the value of

α

is :

Answer: (C)

The integrand has period

π

: on

[k π, (k + 1) π]

set

x = k π + t

, giving

sin^{2} x = sin^{2} t

and

\frac{x}{π} - [\frac{x}{π}] = \frac{t}{π}

. So the integral

= 100 \int_{0}^{π} e^{- t / π} sin^{2} t d t

. With

sin^{2} t = \frac{1 - c o s 2 t}{2}

\int_{0}^{π} e^{- t / π} d t = π (1 - e^{- 1})

and

\int_{0}^{π} e^{- t / π} cos 2 t d t = \frac{π ( 1 - e ^{- 1} )}{1 + 4 π ^{2}}

, so

\int_{0}^{π} e^{- t / π} sin^{2} t d t = \frac{2 π ^{3} ( 1 - e ^{- 1} )}{1 + 4 π ^{2}}

. Thus the value

= \frac{200 π ^{3} ( 1 - e ^{- 1} )}{1 + 4 π ^{2}}

, giving

α = 200 (1 - e^{- 1})

, option (C).

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Single correct option