Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 73338 by mathmax by abdo last updated on 10/Nov/19

calculate ∫_0 ^∞ ((arctan(2cosx))/(3+x^2 ))dx

$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left(\mathrm{2}{cosx}\right)}{\mathrm{3}+{x}^{\mathrm{2}} }{dx} \\ $$

Commented by mathmax by abdo last updated on 11/Nov/19

let I =∫_0 ^∞ ((arctan(2cosx))/(x^2 +3))dx ⇒I =_(x=(√3)t) ∫_0 ^∞ ((arctan(2cos((√3)t)))/(3(t^2 +1)))(√3)dt =(1/(√3))∫_0 ^∞ ((arctan(2cos((√3)t)))/(t^2 +1))dt ⇒2(√3)I =∫_(−∞) ^(+∞) ((arctan(2cos((√3)t))/(t^2 +1))dt let w(z) =((arctan(2cos((√3)z)))/(z^2 +1)) =((arctan(2cos((√3)z)))/((z−i)(z+i))) ∫_(−∞) ^(+∞) w(z)dz =2iπ Res(w,i) Res(w,i) =((arctan(2cos(i(√3))))/(2i)) ⇒∫_(−∞) ^(+∞) w(z)dz =2iπ×((arctan(2cos(i(√3))))/(2i)) =π arctan(2cos(i(√3))) but cos(i(√3))=ch((√3)) =((e^(√3) +e^(−(√3)) )/2) ⇒arctan(2cos(i(√3))) =arctan(e^(√3) +e^(−(√3)) ) ⇒∫_(−∞) ^(+∞) w()dz =π arctan(e^(√3) +e^(−(√3)) ) ⇒ I =(π/(2(√3))) arctan(e^(√3) +e^(−(√3)) ) .

$${let}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left(\mathrm{2}{cosx}\right)}{{x}^{\mathrm{2}} \:+\mathrm{3}}{dx}\:\Rightarrow{I}\:=_{{x}=\sqrt{\mathrm{3}}{t}} \:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left(\mathrm{2}{cos}\left(\sqrt{\mathrm{3}}{t}\right)\right)}{\mathrm{3}\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)}\sqrt{\mathrm{3}}{dt} \\ $$$$=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left(\mathrm{2}{cos}\left(\sqrt{\mathrm{3}}{t}\right)\right)}{{t}^{\mathrm{2}} \:+\mathrm{1}}{dt}\:\Rightarrow\mathrm{2}\sqrt{\mathrm{3}}{I}\:=\int_{−\infty} ^{+\infty} \:\frac{{arctan}\left(\mathrm{2}{cos}\left(\sqrt{\mathrm{3}}{t}\right)\right.}{{t}^{\mathrm{2}} \:+\mathrm{1}}{dt} \\ $$$${let}\:{w}\left({z}\right)\:=\frac{{arctan}\left(\mathrm{2}{cos}\left(\sqrt{\mathrm{3}}{z}\right)\right)}{{z}^{\mathrm{2}} \:+\mathrm{1}}\:=\frac{{arctan}\left(\mathrm{2}{cos}\left(\sqrt{\mathrm{3}}{z}\right)\right)}{\left({z}−{i}\right)\left({z}+{i}\right)} \\ $$$$\int_{−\infty} ^{+\infty} \:{w}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left({w},{i}\right) \\ $$$${Res}\left({w},{i}\right)\:=\frac{{arctan}\left(\mathrm{2}{cos}\left({i}\sqrt{\mathrm{3}}\right)\right)}{\mathrm{2}{i}}\:\Rightarrow\int_{−\infty} ^{+\infty} {w}\left({z}\right){dz}\:=\mathrm{2}{i}\pi×\frac{{arctan}\left(\mathrm{2}{cos}\left({i}\sqrt{\mathrm{3}}\right)\right)}{\mathrm{2}{i}} \\ $$$$=\pi\:{arctan}\left(\mathrm{2}{cos}\left({i}\sqrt{\mathrm{3}}\right)\right)\:\:{but} \\ $$$${cos}\left({i}\sqrt{\mathrm{3}}\right)={ch}\left(\sqrt{\mathrm{3}}\right)\:=\frac{{e}^{\sqrt{\mathrm{3}}} \:+{e}^{−\sqrt{\mathrm{3}}} }{\mathrm{2}}\:\Rightarrow{arctan}\left(\mathrm{2}{cos}\left({i}\sqrt{\mathrm{3}}\right)\right) \\ $$$$={arctan}\left({e}^{\sqrt{\mathrm{3}}} \:+{e}^{−\sqrt{\mathrm{3}}} \right)\:\Rightarrow\int_{−\infty} ^{+\infty} {w}\left(\right){dz}\:=\pi\:{arctan}\left({e}^{\sqrt{\mathrm{3}}} \:+{e}^{−\sqrt{\mathrm{3}}} \right)\:\Rightarrow \\ $$$${I}\:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{3}}}\:{arctan}\left({e}^{\sqrt{\mathrm{3}}} \:+{e}^{−\sqrt{\mathrm{3}}} \right)\:. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com