Is the given expression integrable?

A Quick Solution: Consider the Curve of $y=\sin x$ and and $y=\dfrac{2}{\pi}x$ from $x=0$ to $x=\frac{\pi}{2}$ . Clearly $\sin x$ > $\dfrac{2}{\pi}x$ or $\frac { \sin { x } }{ x } >\frac { 2 }{ \pi }$ Now, Integrating LHS & RHS with limits $x=\frac{\pi}{6}$ to $x=\frac{\pi}{3}$ we get: $\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$ $\displaystyle\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\dfrac{\sin x}{x}dx > \displaystyle\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\dfrac{2}{\pi}dx$ i.e $\displaystyle\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\dfrac{\sin x}{x}dx >\dfrac{1}{3}$ . Now we need to find the upper bound for the integral . As $\sin x < x$ in +ve x -axis. Hence, Doing as above, $\displaystyle\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ \frac { \sin { x } }{ x } } dx<\displaystyle\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ dx }$ or, $\displaystyle\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ \frac { \sin { x } }{ x } } dx <\dfrac{\pi}{6}$
So, $\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$ $\boxed{\dfrac{1}{3}<\displaystyle\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ \frac { \sin { x } }{ x } } dx < \dfrac{\pi}{6}}$

Well, if not interested in doing calculation and to save time in JEE ,it is better to deal with option elimination. As in such types of problems where there are options that can be manipulated through logical thinking and some general concept. As, ( sinx<x<tanx) Therefore, sinx/x will always less than 1. i.e (sinx/x<1). There is only one option where both value is less than 1 is 1/3 and pi/3. I think you all like this way too.

use the fact that sinx>2x/pi and less than x .thus required integral lies in int (2/pi)<I<int(1) with given limits.note: here I is given integral.