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Question Number 59050 by maxmathsup by imad last updated on 04/May/19
calculate ∫_0 ^∞ ((arctan(x^2 ))/(1+x^2 )) dx
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{arctan}\left({x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx}\: \\ $$
Answered by tanmay last updated on 04/May/19
x^2 =tana 2xdx=sec^2 ada dx=((sec^2 ada)/(2(√(tana)) ))da ∫_0 ^(π/2) (a/(1+tana))×((sec^2 a)/(2(√(tana))))da ∫_0 ^(π/2) (a/(cos^2 a())×((cosa)/(sina+cosa)))×((√(cosa))/(2(√(sina))))da ∫_0 ^(π/2) ((ada)/(2(√(sinacosa))))×(1/((sina+cosa))) ∫_0 ^(π/2) ((ada)/( (√(sin2a)) ×(sina+cosa)))=I I=∫_0 ^(π/2) ((((π/2)−a))/( (√(sin2((π/2)−a))) ×(sin((π/2)−a)+cos((π/2)−a)))da I=∫_0 ^(π/2) ((((π/2)−a))/( (√(sin2a)) ×(cosa+sina)))da 2I=∫_0 ^(π/2) ((π/2)/( (√(sin2a)) ×(sina+cosa))) ((4I)/π)=∫_0 ^(π/2) (da/( (√(sin2a)) ×(√(1+sin2a)))) t=tana →dt=sec^2 ada→da=(dt/(1+t^2 )) ((4I)/π)=∫_0 ^∞ (1/( (√((2t)/(1+t^2 )))))×(1/( (√(1+((2t)/(1+t^2 ))))))×(dt/(1+t^2 )) ((4I)/π)=∫_0 ^∞ (dt/( (√(2t)) ×(1+t))) ((4(√2) I)/π)=∫_0 ^∞ (dt/( (√t) (1+t))) k^2 =t→dt=2kdk ((4(√2) I)/π)=∫_0 ^∞ ((2kdk)/(k(1+k^2 ))) ((2(√2))/π)×I=∫_0 ^∞ (dk/(1+k^2 )) ((2(√2))/π)×I=∣tan^(−1) k∣_0 ^∞ I=(π/(2(√2)))×(π/2)→I=(π^2 /(4(√2)))
$${x}^{\mathrm{2}} ={tana} \\ $$$$\mathrm{2}{xdx}={sec}^{\mathrm{2}} {ada} \\ $$$${dx}=\frac{{sec}^{\mathrm{2}} {ada}}{\mathrm{2}\sqrt{{tana}}\:}{da} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{a}}{\mathrm{1}+{tana}}×\frac{{sec}^{\mathrm{2}} {a}}{\mathrm{2}\sqrt{{tana}}}{da} \\ $$$$\left.\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{a}}{{cos}^{\mathrm{2}} {a}\left(\right.}×\frac{{cosa}}{{sina}+{cosa}}\right)×\frac{\sqrt{{cosa}}}{\mathrm{2}\sqrt{{sina}}}{da} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{ada}}{\mathrm{2}\sqrt{{sinacosa}}}×\frac{\mathrm{1}}{\left({sina}+{cosa}\right)} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{ada}}{\:\sqrt{{sin}\mathrm{2}{a}}\:×\left({sina}+{cosa}\right)}={I} \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\left(\frac{\pi}{\mathrm{2}}−{a}\right)}{\:\sqrt{{sin}\mathrm{2}\left(\frac{\pi}{\mathrm{2}}−{a}\right)}\:×\left({sin}\left(\frac{\pi}{\mathrm{2}}−{a}\right)+{cos}\left(\frac{\pi}{\mathrm{2}}−{a}\right)\right.}{da} \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\left(\frac{\pi}{\mathrm{2}}−{a}\right)}{\:\sqrt{{sin}\mathrm{2}{a}}\:×\left({cosa}+{sina}\right)}{da} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\frac{\pi}{\mathrm{2}}}{\:\sqrt{{sin}\mathrm{2}{a}}\:×\left({sina}+{cosa}\right)} \\ $$$$\frac{\mathrm{4}{I}}{\pi}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{da}}{\:\sqrt{{sin}\mathrm{2}{a}}\:×\sqrt{\mathrm{1}+{sin}\mathrm{2}{a}}} \\ $$$${t}={tana}\:\rightarrow{dt}={sec}^{\mathrm{2}} {ada}\rightarrow{da}=\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{4}{I}}{\pi}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\:\sqrt{\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}}×\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{4}{I}}{\pi}=\int_{\mathrm{0}} ^{\infty} \frac{{dt}}{\:\sqrt{\mathrm{2}{t}}\:×\left(\mathrm{1}+{t}\right)} \\ $$$$\frac{\mathrm{4}\sqrt{\mathrm{2}}\:{I}}{\pi}=\int_{\mathrm{0}} ^{\infty} \frac{{dt}}{\:\sqrt{{t}}\:\left(\mathrm{1}+{t}\right)} \\ $$$${k}^{\mathrm{2}} ={t}\rightarrow{dt}=\mathrm{2}{kdk} \\ $$$$\frac{\mathrm{4}\sqrt{\mathrm{2}}\:{I}}{\pi}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2}{kdk}}{{k}\left(\mathrm{1}+{k}^{\mathrm{2}} \right)} \\ $$$$\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\pi}×{I}=\int_{\mathrm{0}} ^{\infty} \frac{{dk}}{\mathrm{1}+{k}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\pi}×{I}=\mid{tan}^{−\mathrm{1}} {k}\mid_{\mathrm{0}} ^{\infty} \\ $$$${I}=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}×\frac{\pi}{\mathrm{2}}\rightarrow\boldsymbol{{I}}=\frac{\pi^{\mathrm{2}} }{\mathrm{4}\sqrt{\mathrm{2}}} \\ $$$$ \\ $$

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