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

Commented by Worm_Tail last updated on 15/Jul/20

$$y={cot}^{-1}\left(x\right)\Rightarrow cot\left(y\right)=x$$$$tan\left(y\right)=\frac{1}{x}\Rightarrow y={tan}^{-1}\frac{1}{x}$$$${cot}^{-1}x={tan}^{-1}\frac{1}{x}$$$$$$$$myplotdownthecommentfromgeogebra$$$$$$

Commented by Worm_Tail last updated on 15/Jul/20

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

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

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

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

Commented by abdomathmax last updated on 16/Jul/20

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

Answered by Worm_Tail last updated on 15/Jul/20

$$f\left(x\right)=\frac{{cot}^{-1}x}{x^4+x^2+1}$$$$f(-x)=\frac{{cot}^{-1}-x}{{(-x)}^4+{(-x)}^2+1}$$$$f(-x)=\frac{-{cot}^{-1}x}{{\left(x\right)}^4+{\left(x\right)}^2+1}=-f\left(x\right)$$$$f\left(x\right)isanoddfunction$$$$\underset{-oo}{\stackrel{oo}{\int }}f\left(x\right)dx=0$$

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

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

$$\mathrm{it}\mathrm{is}\mathrm{not}\mathrm{an}\mathrm{odd}\mathrm{function}$$

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

$${\cot }^{-1}(-\mathrm{x})\ne -{\cot }^{-1}\mathrm{x}$$

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

Commented by Worm_Tail last updated on 15/Jul/20

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