Question Number 103510 by 175mohamed last updated on 15/Jul/20

Commented byWorm_Tail last updated on 15/Jul/20

y=cot^(−1) (x)⇒cot(y)=x  tan(y)=(1/x)⇒y=tan^(−1) (1/x)  cot^(−1) x=tan^(−1) (1/x)    my   plot   down    the  comment  from   geogebra

Commented byWorm_Tail last updated on 15/Jul/20

Commented byPRITHWISH SEN 2 last updated on 15/Jul/20

∵ both are periodic functions but the both do not  have the same period of interval.  the inverse of tan will be in the domain of   −(π/2) to (π/2) and for the cot it is from 0 to π  ∴ you can′t replace cot^(−1)  by tan^(−1) ((1/x)) in  the 0 to −∞ domain.

Commented byPRITHWISH SEN 2 last updated on 15/Jul/20

and this you can see from the graph that   the cot^(−1) x and tan^(−1) (x^(−1) ) graph are not same  in the interval of 0 to −∞.

Commented byabdomathmax last updated on 16/Jul/20

arcotan(x)=y ⇒x =cotany =(1/(tany)) ⇒  (1/x) =tany  ⇒y =arctan((1/x)) ⇒arcotan(x)=arctan((1/x))  ⇒∫_(−∞) ^(+∞)  ((arcotanx)/(x^4  +x^2  +1))dx =∫_(−∞) ^(+∞) ((arctan((1/x)))/(x^4  +x^2 +1))dx =0  because the function under integral is odd

Answered by Worm_Tail last updated on 15/Jul/20

f(x)=((cot^(−1) x)/(x^4 +x^2 +1))  f(−x)=((cot^(−1) −x)/((−x)^4 +(−x)^2 +1))  f(−x)=((−cot^(−1) x)/((x)^4 +(x)^2 +1))=−f(x)  f(x) is an odd function  ∫_(−oo) ^(oo) f(x)dx=0

Commented byPRITHWISH SEN 2 last updated on 15/Jul/20

Commented byPRITHWISH SEN 2 last updated on 15/Jul/20

it is not an odd function

Commented byPRITHWISH SEN 2 last updated on 15/Jul/20

cot^(−1) (−x)≠−cot^(−1) x

Commented byPRITHWISH SEN 2 last updated on 15/Jul/20

Commented byWorm_Tail last updated on 15/Jul/20