Question

The value of ${\int }_0^{2n\pi }[\sin x+\cos x]dx$, is equal to (where $[.]$ denotes greatest integer function)

• A. $-n\pi$
• B. $n\pi$
• C. $-2n\pi$
• D. none of these

Answer 1

The correct option is A $-n\pi$

${\int }_0^{2n\pi }[\sin x+\cos x]dx$

$={\int }_0^{2n\pi }\left[\sqrt{2}\sin (x+\frac{\pi }{4})\right]dx$

The function is periodic with period $2\pi$.

$={\int }_0^{2\pi }\left[\sqrt{2}\sin (x+\frac{\pi }{4})\right]dx$

Hence, first we find the definite integration with limit from $0$ to $2\pi$

The graph of $\sin (x+\frac{\pi }{4})$ is shown in the figure. To find values of $\left[\sqrt{2}\sin (x+\frac{\pi }{4})\right]$, we can draw the lines $y=\frac{1}{\sqrt{2}}$ and $y=-\frac{1}{\sqrt{2}}$

${\int }_0^{2\pi }\left[\sqrt{2}\sin (x+\frac{\pi }{4})\right]dx$

Splitting the limits for greatest integer function,

$={\int }_0^{\frac{\pi }{2}}1.dx+{\int }_{\frac{\pi }{2}}^{3\frac{\pi }{4}}0.dx+{\int }_{\frac{3\pi }{4}}^{\pi }-1.dx+{\int }_{\pi }^{\frac{3\pi }{2}}-2.dx+{\int }_{\frac{3\pi }{2}}^{\frac{7\pi }{4}}-1.dx+{\int }_{\frac{7\pi }{4}}^{2\pi }0.dx$

$=(\frac{\pi }{2}-0)-(\pi -\frac{3\pi }{4})-2(\frac{3\pi }{2}-\pi )-(\frac{7\pi }{4}-\frac{3\pi }{2})$

$=-\pi$

Hence, for limit $0$ to $2n\pi$ the integral will be $n\times -\pi =-n\pi$