JEE MAIN - Mathematics (2009 - No. 8)
$$\int\limits_0^\pi {\left[ {\cot x} \right]dx,} $$ where $$\left[ . \right]$$ denotes the greatest integer function, is equal to:
$$1$$
$$-1$$
$$ - {\pi \over 2}$$
$$ {\pi \over 2}$$
Explanation
Let $$I = \int_0^\pi {\left[ {\cot x} \right]dx\,\,\,\,\,\,...\left( 1 \right)} $$
$$ = \int_0^\pi {\left[ {\cot \left( {\pi - x} \right)} \right]} dx$$
$$ = \int_0^\pi {\left[ { - \cot x} \right]dx\,\,\,\,\,\,...\left( 2 \right)} $$
Adding two values of $$I$$ in $$e{q^n}s\left( 1 \right)\,\,\& \,\,\left( 2 \right),$$
We get $$2I = \int_0^\pi {\left( {\left[ {\cot x} \right] + \left[ { - \cot x} \right]} \right)dx} $$
$$ = \int_0^\pi {\left( { - 1} \right)dx} $$
$$\left[ {} \right.$$ As $$\left[ x \right] + \left[ { - x} \right] = - 1,\,\,if\,\,$$ $$x \notin z$$
and $$\left[ x \right] + \left[ { - x} \right] = 0,\,\,if\,\,x \in z$$ $$\left. {} \right]$$
$$ = \left[ { - x} \right]_0^\pi = - \pi \Rightarrow I = - {\pi \over 2}$$
$$ = \int_0^\pi {\left[ {\cot \left( {\pi - x} \right)} \right]} dx$$
$$ = \int_0^\pi {\left[ { - \cot x} \right]dx\,\,\,\,\,\,...\left( 2 \right)} $$
Adding two values of $$I$$ in $$e{q^n}s\left( 1 \right)\,\,\& \,\,\left( 2 \right),$$
We get $$2I = \int_0^\pi {\left( {\left[ {\cot x} \right] + \left[ { - \cot x} \right]} \right)dx} $$
$$ = \int_0^\pi {\left( { - 1} \right)dx} $$
$$\left[ {} \right.$$ As $$\left[ x \right] + \left[ { - x} \right] = - 1,\,\,if\,\,$$ $$x \notin z$$
and $$\left[ x \right] + \left[ { - x} \right] = 0,\,\,if\,\,x \in z$$ $$\left. {} \right]$$
$$ = \left[ { - x} \right]_0^\pi = - \pi \Rightarrow I = - {\pi \over 2}$$
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