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$\int_0^\pi[\cot x] d x,[\bullet]$ denotes the greatest integer function, is equal to
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2777 Upvotes
Verified Answer
The correct answer is:
$-\frac{\pi}{2}$
$-\frac{\pi}{2}$
$$
\begin{aligned}
& \text { Let } \mathrm{I}=\int_0^\pi[\cot x] d x \\
& =\int_0^\pi[\cot (\pi-x)] d x=\int_0^\pi[-\cot x] d x
\end{aligned}
$$
Adding (1) and (2)
$$
\begin{aligned}
& 2 I=\int_0^\pi[\cot x] d x+\int_0^\pi[-\cot x] d x==\int_0^\pi(-1) d x \quad[\because[x]+[-x]=-1 \text { if } x \notin Z \\
& =[-x]_0^\pi=-\pi \\
& \therefore I=-\frac{\pi}{2}
\end{aligned}
$$
\begin{aligned}
& \text { Let } \mathrm{I}=\int_0^\pi[\cot x] d x \\
& =\int_0^\pi[\cot (\pi-x)] d x=\int_0^\pi[-\cot x] d x
\end{aligned}
$$
Adding (1) and (2)
$$
\begin{aligned}
& 2 I=\int_0^\pi[\cot x] d x+\int_0^\pi[-\cot x] d x==\int_0^\pi(-1) d x \quad[\because[x]+[-x]=-1 \text { if } x \notin Z \\
& =[-x]_0^\pi=-\pi \\
& \therefore I=-\frac{\pi}{2}
\end{aligned}
$$
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