Mathematics - Definite Integration Question with Solution | TestHub

$$\begin{array}{rl}& \text{ Let }\mathrm{I}={\int }_0^{\pi }[\cot x]dx\\ & ={\int }_0^{\pi }[\cot (\pi -x)]dx={\int }_0^{\pi }[-\cot x]dx\end{array}$$ Adding (1) and (2) $$\begin{array}{rl}& 2I={\int }_0^{\pi }[\cot x]dx+{\int }_0^{\pi }[-\cot x]dx=={\int }_0^{\pi }(-1)dx[\because [x]+[-x]=-1\text{ if }x\notin Z\\ & =[-x{]}_0^{\pi }=-\pi \\ & \therefore I=-\frac{\pi }{2}\end{array}$$