Question-103510

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

Commented by Worm_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

$${y}={cot}^{−\mathrm{1}} \left({x}\right)\Rightarrow{cot}\left({y}\right)={x} \\ $$$${tan}\left({y}\right)=\frac{\mathrm{1}}{{x}}\Rightarrow{y}={tan}^{−\mathrm{1}} \frac{\mathrm{1}}{{x}} \\ $$$${cot}^{−\mathrm{1}} {x}={tan}^{−\mathrm{1}} \frac{\mathrm{1}}{{x}} \\ $$$$ \\ $$$${my}\:\:\:{plot}\:\:\:{down}\:\:\:\:{the}\:\:{comment}\:\:{from}\:\:\:{geogebra} \\ $$$$ \\ $$

Commented by Worm_Tail last updated on 15/Jul/20

Commented by PRITHWISH 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.

$$\because\:\mathrm{both}\:\mathrm{are}\:\mathrm{periodic}\:\mathrm{functions}\:\mathrm{but}\:\mathrm{the}\:\mathrm{both}\:\mathrm{do}\:\mathrm{not} \\ $$$$\mathrm{have}\:\mathrm{the}\:\mathrm{same}\:\mathrm{period}\:\mathrm{of}\:\mathrm{interval}. \\ $$$$\mathrm{the}\:\mathrm{inverse}\:\mathrm{of}\:\mathrm{tan}\:\mathrm{will}\:\mathrm{be}\:\mathrm{in}\:\mathrm{the}\:\mathrm{domain}\:\mathrm{of}\: \\ $$$$−\frac{\pi}{\mathrm{2}}\:\mathrm{to}\:\frac{\pi}{\mathrm{2}}\:\mathrm{and}\:\mathrm{for}\:\mathrm{the}\:\mathrm{cot}\:\mathrm{it}\:\mathrm{is}\:\mathrm{from}\:\mathrm{0}\:\mathrm{to}\:\pi \\ $$$$\therefore\:\mathrm{you}\:\mathrm{can}'\mathrm{t}\:\mathrm{replace}\:\mathrm{cot}^{−\mathrm{1}} \:\mathrm{by}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{x}}\right)\:\mathrm{in} \\ $$$$\mathrm{the}\:\mathrm{0}\:\mathrm{to}\:−\infty\:\mathrm{domain}. \\ $$

Commented by PRITHWISH 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 −∞.

$$\mathrm{and}\:\mathrm{this}\:\mathrm{you}\:\mathrm{can}\:\mathrm{see}\:\mathrm{from}\:\mathrm{the}\:\mathrm{graph}\:\mathrm{that}\: \\ $$$$\mathrm{the}\:\mathrm{cot}^{−\mathrm{1}} \mathrm{x}\:\mathrm{and}\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}^{−\mathrm{1}} \right)\:\mathrm{graph}\:\mathrm{are}\:\mathrm{not}\:\mathrm{same} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{interval}\:\mathrm{of}\:\mathrm{0}\:\mathrm{to}\:−\infty. \\ $$

Commented by abdomathmax 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

$$\mathrm{arcotan}\left(\mathrm{x}\right)=\mathrm{y}\:\Rightarrow\mathrm{x}\:=\mathrm{cotany}\:=\frac{\mathrm{1}}{\mathrm{tany}}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{x}}\:=\mathrm{tany}\:\:\Rightarrow\mathrm{y}\:=\mathrm{arctan}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)\:\Rightarrow\mathrm{arcotan}\left(\mathrm{x}\right)=\mathrm{arctan}\left(\frac{\mathrm{1}}{\mathrm{x}}\right) \\ $$$$\Rightarrow\int_{−\infty} ^{+\infty} \:\frac{\mathrm{arcotanx}}{\mathrm{x}^{\mathrm{4}} \:+\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dx}\:=\int_{−\infty} ^{+\infty} \frac{\mathrm{arctan}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)}{\mathrm{x}^{\mathrm{4}} \:+\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{dx}\:=\mathrm{0} \\ $$$$\mathrm{because}\:\mathrm{the}\:\mathrm{function}\:\mathrm{under}\:\mathrm{integral}\:\mathrm{is}\:\mathrm{odd} \\ $$

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

$${f}\left({x}\right)=\frac{{cot}^{−\mathrm{1}} {x}}{{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$${f}\left(−{x}\right)=\frac{{cot}^{−\mathrm{1}} −{x}}{\left(−{x}\right)^{\mathrm{4}} +\left(−{x}\right)^{\mathrm{2}} +\mathrm{1}} \\ $$$${f}\left(−{x}\right)=\frac{−{cot}^{−\mathrm{1}} {x}}{\left({x}\right)^{\mathrm{4}} +\left({x}\right)^{\mathrm{2}} +\mathrm{1}}=−{f}\left({x}\right) \\ $$$${f}\left({x}\right)\:{is}\:{an}\:{odd}\:{function} \\ $$$$\underset{−{oo}} {\overset{{oo}} {\int}}{f}\left({x}\right){dx}=\mathrm{0} \\ $$

Commented by PRITHWISH SEN 2 last updated on 15/Jul/20