Integration Questions

Question Number 77086 by Dah Solu Tion last updated on 03/Jan/20

$$\int_{\mathrm{0}} ^{\mathrm{1}} {xtan}^{−\mathrm{1}} {xdx} \\$$$$\\$$

Commented by Tony Lin last updated on 03/Jan/20

$$\int{xtan}^{−\mathrm{1}} {xdx}\:\:\left(\int{f}\:'{g}={fg}−\int{fg}'\right) \\$$$$=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}{tan}^{−\mathrm{1}} {x}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\$$$$=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}{tan}^{−\mathrm{1}} {x}−\frac{\mathrm{1}}{\mathrm{2}}{x}+\frac{\mathrm{1}}{\mathrm{2}}{tan}^{−\mathrm{1}} {x}+{c} \\$$$$=\frac{\left({x}^{\mathrm{2}} +\mathrm{1}\right){tan}^{−\mathrm{1}} {x}−{x}}{\mathrm{2}}+{c} \\$$$$\int_{\mathrm{0}} ^{\mathrm{1}} {xtan}^{−\mathrm{1}} {xdx} \\$$$$=\frac{\mathrm{2}×\frac{\pi}{\mathrm{4}}−\mathrm{1}}{\mathrm{2}} \\$$$$=\frac{\pi−\mathrm{2}}{\mathrm{4}} \\$$

Commented by turbo msup by abdo last updated on 03/Jan/20

$${let}\:{A}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:{xarctanxdx}\:{by}\:{parts} \\$$$${A}\:=\left[\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:{arctanx}\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} }{\mathrm{2}}\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\pi}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{x}^{\mathrm{2}} +\mathrm{1}−\mathrm{1}}{{x}^{\mathrm{2}} \:+\mathrm{1}}{dx} \\$$$$=\frac{\pi}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\left[{arctanx}\right]_{\mathrm{0}} ^{\mathrm{1}} \\$$$$=\frac{\pi}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{2}}+\:\frac{\pi}{\mathrm{8}}\:=\frac{\pi}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}} \\$$

Answered by john santu last updated on 03/Jan/20

$${using}\:{integration}\:{by}\:{part}\: \\$$$${u}\:=\:\mathrm{tan}^{−\mathrm{1}} \left({x}\right)\:\Rightarrow{du}\:=\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\:{dx} \\$$$${dv}\:=\:{x}\:{dx}\:\Rightarrow\:{v}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} \\$$$${I}\:=\:{u}.{v}\:−\:\int\:{v}\:{du}\: \\$$$$=\:\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} \mathrm{tan}^{−\mathrm{1}} \left({x}\right)−\int\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{1}}\right){dx} \\$$$$=\:\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} \mathrm{tan}^{−\mathrm{1}} \left({x}\right)\:−\frac{\mathrm{1}}{\mathrm{2}}\left(\int\mathrm{1}−\mathrm{sec}\:^{\mathrm{2}} {t}\right){dt}\:,\:\left({x}=\mathrm{tan}\:{t}\right) \\$$$$=\:\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} \mathrm{tan}^{−\mathrm{1}} \left({x}\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \left({x}\right)+{x}\:\bigstar \\$$$${now}\:{please}\:{substuti}\:{x}\:=\:\mathrm{1}\:{and}\:{x}\:=\:\mathrm{0}.\:\bigstar \\$$