Question

${\int }_0^{\pi }\left[\cot x\right]dx$, where [.] denotes the greatest integer function, is equal to:

• A. $1$
• B. $-1$
• C. $-\pi /2$
• D. $\pi /2$

Answer 1

Answer: (C)

$-\pi /2$

$I={\int }_0^{\pi }\left[cotx\right]dx$ ...(1)
Using property ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$
$I={\int }_0^{\pi }\left[cot(\pi -x)\right]dx={\int }_0^{\pi }[-cotx]dx$ ...(2)
Adding (1) and (2), we get
$2I={\int }_0^{\pi }(\left[cotx\right]+[-cotx])dx$
As $\left[x\right]+[-x]=-1$
$2I={\int }_0^{\pi }(-1)dx=-{\left[x\right]}_0^{\pi }=-\pi I=-\frac{\pi }{2}$