JEE Main 2022 — Definite Integration Question with Solution

We know,

$$\left[\frac{x}{2}\right]$$ is discontinuous at 1, 2, 3, 4 ........

$$\therefore$$ [ x ] is discontinuous at 2, 4, 6, 8 .....

In between 0 to 5 it is discontinuous at 2 and 4.

Break the integration into 3 parts

(1) 0 to 2

(2) 2 to 4

(3) 4 to 5

$$\therefore$$ $$\underset{0}{\overset{5}{\int }}\cos \left(\pi \left(x-\left[\frac{x}{2}\right]\right)\right)dx$$

$$=\underset{0}{\overset{2}{\int }}\cos \left(\pi (x-0)\right)dx+\underset{2}{\overset{4}{\int }}\cos \left(\pi (x-1)\right)dx+\underset{4}{\overset{5}{\int }}\cos \left(\pi (x-2)\right)dx$$

$$=\underset{0}{\overset{2}{\int }}\cos \pi xdx+\underset{2}{\overset{4}{\int }}\cos (\pi x-\pi )dx+\underset{4}{\overset{5}{\int }}\cos (\pi x-2\pi )dx$$

$$=\underset{0}{\overset{2}{\int }}\cos \pi dx-\underset{2}{\overset{4}{\int }}\cos \pi xdx+\underset{4}{\overset{5}{\int }}\cos \pi xdx$$

$$={\left[\frac{\sin \pi x}{\pi }\right]}_0^2-{\left[\frac{\sin \pi x}{\pi }\right]}_2^4+{\left[\frac{\sin \pi x}{\pi }\right]}_4^5$$

$$=0-0+0$$

$$=0$$