Integration Questions

Question Number 70718 by oyemi kemewari last updated on 07/Oct/19

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{tan}^{−\mathrm{1}} \mathrm{x}}{\mathrm{1}+\mathrm{x}}\mathrm{dx} \\$$

Commented by mathmax by abdo last updated on 07/Oct/19

$${let}\:{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{arctan}\left({x}\right)}{\mathrm{1}+{x}}{dx}\:\:{and}\:{f}\left({t}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{arctan}\left({tx}\right)}{\mathrm{1}+{x}}{dx}\:\:{with}\:{t}\geqslant\mathrm{0} \\$$$${I}={f}\left(\mathrm{1}\right)\:\:{and}\:{f}^{'} \left({t}\right)=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{x}}{\left(\mathrm{1}+{x}^{\mathrm{2}} {t}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}\right)}{dx} \\$$$$=_{{xt}={u}} \:\:\:\:\:\:\int_{\mathrm{0}} ^{{t}} \:\:\:\frac{{u}}{{t}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\mathrm{1}+\frac{{u}}{{t}}\right)}\:\frac{{du}}{{t}}\:=\int_{\mathrm{0}} ^{{t}} \:\:\:\frac{{u}}{{t}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left({t}+{u}\right)}{du} \\$$$$=\frac{\mathrm{1}}{{t}}\:\int_{\mathrm{0}} ^{{t}} \:\:\frac{{udu}}{\left({u}+{t}\right)\left({u}^{\mathrm{2}} +\mathrm{1}\right)}\:\:{let}\:{decompose}\:{F}\left({u}\right)=\frac{{u}}{\left({u}+{t}\right)\left({u}^{\mathrm{2}} +\mathrm{1}\right)} \\$$$${F}\left({u}\right)=\frac{{a}}{{u}+{t}}\:+\frac{{bu}\:+{c}}{{u}^{\mathrm{2}} \:+\mathrm{1}} \\$$$${a}={lim}_{{u}\rightarrow−{t}} \:\:\:\left({u}+{t}\right){F}\left({u}\right)=\frac{−{t}}{{t}^{\mathrm{2}} +\mathrm{1}} \\$$$${lim}_{{u}\rightarrow+\infty} {uF}\left({u}\right)=\mathrm{0}\:={a}+{b}\:\Rightarrow{b}=\frac{{t}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\$$$${F}\left({u}\right)=\frac{−{t}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left({u}+{t}\right)}\:+\frac{\frac{{tu}}{{t}^{\mathrm{2}} +\mathrm{1}}+{c}}{{u}^{\mathrm{2}} \:+\mathrm{1}} \\$$$${F}\left(\mathrm{0}\right)=\mathrm{0}=\frac{−{t}}{{t}\left({t}^{\mathrm{2}} +\mathrm{1}\right)}\:+{c}\:=−\frac{\mathrm{1}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:+{c}\:\Rightarrow{c}=\frac{\mathrm{1}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\$$$${F}\left({u}\right)=\frac{\mathrm{1}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\left\{\frac{−{t}}{{u}+{t}}\:+\frac{{tu}\:+\mathrm{1}}{{u}^{\mathrm{2}} \:+\mathrm{1}}\right\}\:\Rightarrow \\$$$${f}^{'} \left({t}\right)=\int_{\mathrm{0}} ^{{t}} {F}\left({u}\right){du}\:=\frac{−{t}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:\int_{\mathrm{0}} ^{{t}} \:\:\frac{{du}}{{u}+{t}}\:+\frac{\mathrm{1}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\frac{{t}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{{t}} \frac{\mathrm{2}{u}}{{u}^{\mathrm{2}} \:+\mathrm{1}}{du}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}\int_{\mathrm{0}} ^{{t}} \:\frac{{du}}{{u}^{\mathrm{2}} \:+\mathrm{1}} \\$$$$=−\frac{{t}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\left[{ln}\left({u}+{t}\right)\right]_{\mathrm{0}} ^{{t}} \:\:+\frac{{t}}{\mathrm{2}\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)}\left[{ln}\left({u}^{\mathrm{2}} +\mathrm{1}\right)\right]_{\mathrm{0}} ^{{t}} \:+\frac{{arctan}\left({t}\right)}{{t}^{\mathrm{2}} \:+\mathrm{1}} \\$$$$=\frac{−{tln}\left(\mathrm{2}\right)}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{{tln}\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{2}\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)}\:+\frac{{arctan}\left({t}\right)}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\$$$${f}\left({t}\right)=−{ln}\left(\mathrm{2}\right)\int_{\mathrm{0}} ^{{t}} \:\:\frac{{x}}{{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}\:+\int_{\mathrm{0}} ^{{t}} \:\:\frac{{xln}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)}{\mathrm{2}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)}{dx}\:\:+\int_{\mathrm{0}} ^{{t}} \:\frac{{arctan}\left({x}\right)}{{x}^{\mathrm{2}} \:+\mathrm{1}}{dx} \\$$$$+{c}\:.....{be}\:{continued}... \\$$

Answered by mind is power last updated on 07/Oct/19

$$\mathrm{x}=\mathrm{tg}\left(\mathrm{u}\right) \\$$$$\Rightarrow{dx}=\frac{\mathrm{du}}{\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{u}\right)} \\$$$$\mathrm{I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}\right)}{\mathrm{1}+\mathrm{x}}\mathrm{dd}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{u}}{\left(\mathrm{1}+{tg}\left({u}\right)\right)}.\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \left({u}\right)}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{udu}}{\left({sin}\left({u}\right)+{cos}\left({u}\right)\right)\left({cos}\left({u}\right)\right)} \\$$$${cos}\left({u}\right)+{sin}\left({u}\right)=\sqrt{\mathrm{2}}\left({cos}\left({u}−\frac{\pi}{\mathrm{4}}\right)\right) \\$$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{udu}}{\sqrt{\mathrm{2}}{cos}\left({u}−\frac{\pi}{\mathrm{4}}\right){cos}\left({u}\right)} \\$$$${let}\:{f}\left({u}\right)=\frac{{u}}{\sqrt{\mathrm{2}}{cos}\left({u}−\frac{\pi}{\mathrm{4}}\right){cos}\left({u}\right)} \\$$$${i}\:{will}\:{use}\:{this}\:{in}\:{the}\:{next}\:{step}\int_{{a}} ^{{b}} {f}\left({x}\right){dx}=\int_{{a}} ^{{b}} {f}\left({a}+{b}−{x}\right){dx} \\$$$$\Rightarrow{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\frac{\pi}{\mathrm{4}}−{u}}{\sqrt{\mathrm{2}}{cos}\left(\frac{\pi}{\mathrm{4}}−{u}−\frac{\pi}{\mathrm{4}}\right).{cos}\left(\frac{\pi}{\mathrm{4}}−{u}\right)}{du} \\$$$${I}=\int_{\mathrm{0}} ^{{u}} \frac{−{udu}}{\sqrt{\mathrm{2}}{cos}\left({u}\right){cos}\left(\frac{\pi}{\mathrm{4}}−{u}\right)}+\frac{\pi}{\mathrm{4}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{du}}{\sqrt{\mathrm{2}}{cos}\left(\frac{\pi}{\mathrm{4}}−{u}\right){cos}\left({u}\right)} \\$$$$\Rightarrow{I}=−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}.{I}+\frac{\pi}{\mathrm{4}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{du}}{\left({cos}\left({u}\right)+{sin}\left({u}\right)\right){cos}\left({u}\right)} \\$$$$\Rightarrow{I}=\frac{\sqrt{\mathrm{2}}}{\sqrt{\mathrm{2}+}\mathrm{1}}.\frac{\pi}{\mathrm{4}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{1}.{du}}{\left({cos}\left({u}\right)+{sin}\left({u}\right)\right){cos}\left({u}\right)} \\$$$$\mathrm{1}=\left({cos}\left({u}\right)+{sin}\left({u}\right)\right)^{\mathrm{2}} −\mathrm{2}{sin}\left({u}\right){cos}\left({u}\right) \\$$$${A}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{1}.{du}}{\left({cos}\left({u}\right)+{sin}\left({u}\right)\right){cos}\left({u}\right)}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\left({cos}\left({u}\right)+{sin}\left({u}\right)\right)^{\mathrm{2}} −\mathrm{2}{cos}\left({u}\right){sin}\left({u}\right)}{\left({sin}\left({u}\right)+{cos}\left({u}\right)\right){cos}\left({u}\right)} \\$$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{cos}\left({u}\right)+{sin}\left({u}\right)}{{cos}\left({u}\right)}{du}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{2}{sin}\left({u}\right)}{{cos}\left({u}\right)+{sin}\left({u}\right)}{du} \\$$$${sin}\left({u}\right)=−\left({cos}\left({u}\right)−{sin}\left({u}\right)+\left({sin}\left({u}\right)+{cos}\left({u}\right)\right)\right. \\$$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left(\mathrm{1}+{tg}\left({u}\right)\right){du}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{−\left({cos}\left({u}\right)−{sin}\left({u}\right)\right)+\left({sin}\left({u}\right)+{cos}\left({u}\right)\right){du}}{{cos}\left({u}\right)+{sin}\left({u}\right)} \\$$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left(\mathrm{1}+{tg}\left({u}\right)\right){du}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{cos}\left({u}\right)−{sin}\left({u}\right)}{{cos}\left({u}\right)+{sin}\left({u}\right)}{du}−\int\mathrm{1}{du} \\$$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {tg}\left({u}\right){du}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{d}\left({sin}\left({u}\right)+{cos}\left({u}\right)\right)}{{cos}\left({u}\right)+{sin}\left({u}\right)} \\$$$$=_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left[−{ln}\left({cos}\left({u}\right)\right)+{ln}\left({cos}\left({u}\right)+{sin}\left({u}\right)\right)\right] \\$$$$=−{ln}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)+{ln}\left(\sqrt{\mathrm{2}}\right)={ln}\left(\mathrm{2}\right) \\$$$${I}=\frac{\sqrt{\mathrm{2}}}{\sqrt{\mathrm{2}}+\mathrm{1}}.\frac{\pi}{\mathrm{4}}.{A}=\frac{\pi.{ln}\left(\mathrm{2}\right)\sqrt{\mathrm{2}}}{\mathrm{4}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)} \\$$$$\\$$$$\\$$

Commented by oyemi kemewari last updated on 07/Oct/19

thank you sir

Commented by mind is power last updated on 07/Oct/19

$${y}'{re}\:{welcom} \\$$

Commented by oyemi kemewari last updated on 08/Oct/19

thank you sir