Question Number 66695 by mathmax by abdo last updated on 18/Aug/19
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left({arctanx}\right)}{\mathrm{4}+{x}^{\mathrm{2}} }{dx} \\ $$
Commented by ~ À ® @ 237 ~ last updated on 18/Aug/19
$$\:{Let}\:{named}\:{it}\:{J}\: \\ $$$${As}\:\mathrm{1}+{tan}^{\mathrm{2}} {t}=\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {t}}\:\:{we}\:{have}\:\:{cos}\left({arctanx}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:} \\ $$$${Now}\:\:{J}=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left(\mathrm{4}+{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\: \\ $$$${let}\:{state}\:{x}={shu}\:\Rightarrow\:{du}=\frac{{dx}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:} \\ $$$${J}=\:\int_{\mathrm{0}} ^{\infty} \frac{\:{du}}{\mathrm{4}+{sh}^{\mathrm{2}} {u}}\:=\:\int_{\mathrm{0}} ^{\infty} \:\frac{\frac{\mathrm{1}}{{ch}^{\mathrm{2}} {u}}}{\frac{\mathrm{4}}{{ch}^{\mathrm{2}} {u}}\:+\:{th}^{\mathrm{2}} {u}}\:{du}\: \\ $$$${As}\:\:\:\frac{\mathrm{1}}{{ch}^{\mathrm{2}} {u}}\:=\mathrm{1}−{th}^{\mathrm{2}} {u}\:\:{and}\:\:\frac{{d}}{{du}}\:\left({thu}\right)=\frac{\mathrm{1}}{{ch}^{\mathrm{2}} {u}}\:\:\:\:\:{we}\:{have} \\ $$$${J}=\:\int_{\mathrm{0}} ^{\infty} \:\frac{{d}\left({thu}\right)}{\mathrm{4}\:−\mathrm{3}{th}^{\mathrm{2}} {u}}\:=\:\frac{\mathrm{1}}{\mathrm{4}}.\:\int_{\mathrm{0}} ^{\infty} \:\frac{{d}\left({thu}\right)}{\mathrm{1}−\left(\frac{\sqrt{\mathrm{3}}\:}{\mathrm{2}}\:{thu}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\:.\int_{\mathrm{0}} ^{\infty} \:\frac{\:{d}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{thu}\right)}{\mathrm{1}−\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{thu}\right)^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\:\left[{argth}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{thu}\right)\right]_{\mathrm{0}} ^{\infty} =\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\:{argth}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$${finally} \\ $$$${J}=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\:.\:\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{1}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}{\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\right)=\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{3}}\:}.\:{ln}\left(\frac{\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}}}\:\right)=\frac{{ln}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)}{\:\sqrt{\mathrm{3}}}\:−\frac{{ln}\mathrm{2}}{\mathrm{2}\sqrt{\mathrm{3}}}\: \\ $$
Commented by mathmax by abdo last updated on 19/Aug/19
$${thanks}\:{sir}. \\ $$
Commented by mathmax by abdo last updated on 19/Aug/19
$${let}\:{I}=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left({arctanx}\right)}{\mathrm{4}+{x}^{\mathrm{2}} }{dx}\:\:{we}\:{know}\:{that}\:{cos}\left({arctanx}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$$\Rightarrow\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{4}\right)\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:\:{changement}\:\:{x}={sht}\:{give} \\ $$$${I}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{ch}\left({t}\right)}{\left({sh}^{\mathrm{2}} {t}\:+\mathrm{4}\right){ch}\left({t}\right)}{dt}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dt}}{\mathrm{4}+\frac{{ch}\left(\mathrm{2}{t}\right)−\mathrm{1}}{\mathrm{2}}}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{2}{dt}}{\mathrm{8}+{ch}\left(\mathrm{2}{t}\right)−\mathrm{1}} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{2}{dt}}{\mathrm{7}+{ch}\left(\mathrm{2}{t}\right)}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{2}{dt}}{\mathrm{7}+\frac{{e}^{\mathrm{2}{t}} \:+{e}^{−\mathrm{2}{t}} }{\mathrm{2}}}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{4}{dt}}{\mathrm{14}+{e}^{\mathrm{2}{t}} \:+{e}^{−\mathrm{2}{t}} } \\ $$$$=_{{e}^{\mathrm{2}{t}} ={u}} \:\:\:\int_{\mathrm{1}} ^{+\infty} \:\:\:\:\frac{\mathrm{4}}{\mathrm{14}\:+{u}+{u}^{−\mathrm{1}} }\:\frac{{du}}{\mathrm{2}{u}}\:=\:\int_{\mathrm{1}} ^{+\infty} \:\:\frac{\mathrm{2}{du}}{\mathrm{14}{u}\:+{u}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$=\int_{\mathrm{1}} ^{+\infty} \:\frac{\mathrm{2}{du}}{{u}^{\mathrm{2}} \:+\mathrm{14}{u}\:+\mathrm{1}} \\ $$$${u}^{\mathrm{2}} \:+\mathrm{14}{u}+\mathrm{1}=\mathrm{0}\rightarrow\Delta^{'} =\mathrm{7}^{\mathrm{2}} −\mathrm{1}\:=\mathrm{48}\:\Rightarrow{u}_{\mathrm{1}} =−\mathrm{7}+\sqrt{\mathrm{48}}=−\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}} \\ $$$${u}_{\mathrm{2}} =−\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}\:\Rightarrow{I}\:=\int_{\mathrm{1}} ^{+\infty} \:\:\frac{\mathrm{2}{du}}{\left({u}−{u}_{\mathrm{1}} \right)\left({u}−{u}_{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{2}}{\mathrm{8}\sqrt{\mathrm{3}}}\int_{\mathrm{1}} ^{+\infty} \:\left\{\frac{\mathrm{1}}{{u}−{u}_{\mathrm{1}} }−\frac{\mathrm{1}}{{u}−{u}_{\mathrm{2}} }\right\}{du}\:=\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{3}}}\left[{ln}\mid\frac{{u}−{u}_{\mathrm{1}} }{{u}−{u}_{\mathrm{2}} }\mid\right]_{\mathrm{1}} ^{+\infty} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{3}}}\left\{−{ln}\mid\frac{\mathrm{1}−{u}_{\mathrm{1}} }{\mathrm{1}−{u}_{\mathrm{2}} }\mid\right\}\:=\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{3}}}\left\{{ln}\mid\frac{\mathrm{1}+\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{1}+\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}}\mid\right\}\:=\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{3}}}{ln}\mid\frac{\mathrm{8}+\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{8}−\mathrm{4}\sqrt{\mathrm{3}}}\mid \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{3}}}{ln}\left(\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{2}−\sqrt{\mathrm{3}}}\right)\:. \\ $$
Answered by mind is power last updated on 18/Aug/19
$${cos}\left({arctg}\left({x}\right)\right)={cos}\left({x}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{tg}^{\mathrm{2}} \left({x}\right)}}==>\frac{\mathrm{1}}{\:\sqrt{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}}={cos}\left({arctg}\left({x}\right)\right).{x}\geqslant\mathrm{0} \\ $$$$=\int_{\mathrm{0}} ^{+\infty} \frac{{dx}}{\:\sqrt{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\left(\mathrm{4}+{x}^{\mathrm{2}} \right)} \\ $$$${x}={sh}\left({t}\right) \\ $$$${dx}={ch}\left({t}\right) \\ $$$$==>\int_{\mathrm{0}} ^{+\infty} \frac{{ch}\left({t}\right){dt}}{{ch}\left({t}\right)\left(\mathrm{4}+{sh}^{\mathrm{2}} \left({t}\right)\right)} \\ $$$${withe}\:{e}^{{t}} ={w}=\geqslant{dt}=\frac{\mathrm{1}}{{w}}{dw} \\ $$$$==>\int_{\mathrm{1}} ^{+\infty} \frac{{dw}}{{w}\left(\mathrm{4}+\frac{{w}^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}{w}^{\mathrm{2}} }−\mathrm{1}\right)}=\int_{−\infty} ^{+\infty} \frac{\mathrm{4}{wdw}}{\left({w}^{\mathrm{4}} +\mathrm{12}{w}^{\mathrm{2}} +\mathrm{1}\right)}=\int\frac{\mathrm{4}{w}}{\left({w}^{\mathrm{2}} +\mathrm{6}−\sqrt{\mathrm{35}}\right)\left({w}^{\mathrm{2}} +\mathrm{6}+\sqrt{\mathrm{35}}\right)} \\ $$$$\frac{\mathrm{4}{w}}{\left({w}^{\mathrm{2}} +\mathrm{6}−\sqrt{\left.\mathrm{35}\right)}\left({w}^{\mathrm{2}} +\mathrm{6}+\sqrt{\mathrm{35}}\right)\right.}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{35}}}\left(\frac{{w}}{{w}^{\mathrm{2}} +\mathrm{6}−\sqrt{\mathrm{35}}}−\frac{{w}}{{w}^{\mathrm{2}} +\mathrm{6}+\sqrt{\mathrm{35}}}\right) \\ $$$$==>\int\frac{\mathrm{4}{wdw}}{{w}^{\mathrm{4}} −\mathrm{12}{w}^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{35}}}{ln}\left(\frac{{w}^{\mathrm{2}} +\mathrm{6}−\sqrt{\mathrm{35}}}{{w}^{\mathrm{2}} +\mathrm{6}+\sqrt{\mathrm{35}}}\right)+{c} \\ $$$$==>\int_{\mathrm{1}} ^{+\infty} \frac{\mathrm{4}{wdw}}{{w}^{\mathrm{4}} −\mathrm{12}{w}^{\mathrm{2}} +\mathrm{1}}={li}\underset{{x}\rightarrow\infty} {{m}}\left[\frac{\mathrm{2}}{\:\sqrt{\mathrm{35}}}\mathrm{ln}\:\left(\frac{{w}^{\mathrm{2}} +\mathrm{6}−\sqrt{\mathrm{35}}}{{w}^{\mathrm{2}} +\mathrm{6}+\sqrt{\mathrm{35}}}\right)\right]_{\mathrm{1}} ^{{x}} \\ $$$$={lim}_{{x}\rightarrow+\infty} \frac{\mathrm{2}}{\:\sqrt{\mathrm{35}}}{ln}\left(\frac{{x}^{\mathrm{2}} +\mathrm{6}−\sqrt{\mathrm{35}}}{{x}^{\mathrm{2}} +\mathrm{6}+\sqrt{\mathrm{35}}}\right)+\frac{\mathrm{2}}{\:\sqrt{\mathrm{35}}}{ln}\left(\frac{\mathrm{7}+\sqrt{\mathrm{35}}}{\mathrm{7}−\sqrt{\mathrm{35}}}\right)=\frac{\mathrm{2}}{\:\sqrt{\mathrm{35}}}{ln}\left(\frac{\mathrm{7}+\sqrt{\mathrm{35}}}{\mathrm{7}−\sqrt{\mathrm{35}}}\right) \\ $$$$ \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 19/Aug/19
$${thanks}\:{sir}. \\ $$