$$\int_0^5 {\cos \left( {\pi \left( {x - \left[ {{x \over 2}} \right]} \right)} \right)dx} $$,

where [t] denotes greatest integer less than or equal to t, is equal to:

<p>We know,</p> <p>$\left[ {{x \over 2}} \right]$ is discontinuous at 1, 2, 3, 4 ........</p> <p>$\therefore$ [ x ] is discontinuous at 2, 4, 6, 8 .....</p> <p>In between 0 to 5 it is discontinuous at 2 and 4.</p> <p>Break the integration into 3 parts</p> <p>(1) 0 to 2</p> <p>(2) 2 to 4</p> <p>(3) 4 to 5</p> <p>$\therefore$ $$\int\limits_0^5 {\cos \left( {\pi \left( {x - \left[ {{x \over 2}} \right]} \right)} \right)dx} $$</p> <p>$$ = \int\limits_0^2 {\cos \left( {\pi (x - 0)} \right)dx + \int\limits_2^4 {\cos \left( {\pi (x - 1)} \right)dx + \int\limits_4^5 {\cos \left( {\pi (x - 2)} \right)dx} } } $$</p> <p>$$ = \int\limits_0^2 {\cos \pi x\,dx + \int\limits_2^4 {\cos (\pi x - \pi )dx + \int\limits_4^5 {\cos (\pi x - 2\pi )dx} } } $$</p> <p>$$ = \int\limits_0^2 {\cos \pi dx - \int\limits_2^4 {\cos \pi x\,dx + \int\limits_4^5 {\cos \pi x\,dx} } } $$</p> <p>$$ = \left[ {{{\sin \pi x} \over \pi }} \right]_0^2 - \left[ {{{\sin \pi x} \over \pi }} \right]_2^4 + \left[ {{{\sin \pi x} \over \pi }} \right]_4^5$$</p> <p>$= 0 - 0 + 0$</p> <p>$= 0$</p>