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Question Number 34562 by math khazana by abdo last updated on 08/May/18
find the value of ∫_0 ^1 ((arctanx)/((1+x^2 )^2 )) dx
$${find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{arctanx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dx} \\ $$
Commented by math khazana by abdo last updated on 08/May/18
we have proved that ∫ ((arctanx)/((1+x^2 )^2 ))dx=(1/4) (arctanx)^2 +(1/4)sin(2arctanx)+k⇒ ∫_0 ^1 ((arctanx)/((1+x^2 )^2 ))dx=[(1/4)(arctanx)^2 +(1/4)sin(2arctanx)]_0 ^1 =(1/4)((π/4))^2 +(1/4)sin((π/2)) = (π^2 /(64)) +(1/4) .
$${we}\:{have}\:{proved}\:{that}\:\: \\ $$$$\int\:\:\:\:\:\frac{{arctanx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}=\frac{\mathrm{1}}{\mathrm{4}}\:\left({arctanx}\right)^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{2}{arctanx}\right)+{k}\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{arctanx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}=\left[\frac{\mathrm{1}}{\mathrm{4}}\left({arctanx}\right)^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{2}{arctanx}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\pi}{\mathrm{4}}\right)^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\frac{\pi}{\mathrm{2}}\right)\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{64}}\:\:+\frac{\mathrm{1}}{\mathrm{4}}\:\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 08/May/18
t=tan^(−1) x,(dt/dx)=(1/(1+x^2 )) =∫_0 ^(Π/4) (t/((1+tan^2 t)^2 )).(1+tan^2 t)dt =∫_0 ^(Π/4) t.cos^2 tdt =∫_0 ^(Π/4) t(1+cos2t)/2dt =(1/2)∫_0 ^(Π/4) t dt +(1/2)∫_0 ^(Π/4) tcos2t dt =(1/2)∣(t^2 /2)∣_0 ^(Π/4) +I_2 =(1/4).(Π^2 /(16))=(Π^2 /(64)) I_2 =(1/2)∫_0 ^(Π/4) tcos2t dt (1/2)∣((tsin2t)/2)∣_0 ^(Π/4) −I_3 ∫tcos2t dt =t((sin2t)/2)−∫{(dt/(dt )).∫cos2t dt}dt =t((sin2t)/2)−∫((sin2t)/2)dt =t((sin2t)/2)+((cos2t)/4) so I_3 =∣((tsin2t)/2)+((cos2t)/4)∣_0 ^(Π/4)
$${t}={tan}^{−\mathrm{1}} {x},\frac{{dt}}{{dx}}=\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$=\underset{\mathrm{0}} {\overset{\Pi/\mathrm{4}} {\int}}\frac{{t}}{\left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\right)^{\mathrm{2}} }.\left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\right){dt} \\ $$$$=\underset{\mathrm{0}} {\overset{\Pi/\mathrm{4}} {\int}}{t}.{cos}^{\mathrm{2}} {tdt} \\ $$$$=\underset{\mathrm{0}} {\overset{\Pi/\mathrm{4}} {\int}}{t}\left(\mathrm{1}+{cos}\mathrm{2}{t}\right)/\mathrm{2}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\Pi/\mathrm{4}} {\int}}{t}\:{dt}\:+\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\Pi/\mathrm{4}} {\int}}{tcos}\mathrm{2}{t}\:{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mid\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\mid_{\mathrm{0}} ^{\Pi/\mathrm{4}} +{I}_{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}.\frac{\prod^{\mathrm{2}} }{\mathrm{16}}=\frac{\prod^{\mathrm{2}} }{\mathrm{64}} \\ $$$${I}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\Pi/\mathrm{4}} {tcos}\mathrm{2}{t}\:{dt} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mid\frac{{tsin}\mathrm{2}{t}}{\mathrm{2}}\mid_{\mathrm{0}} ^{\Pi/\mathrm{4}} −{I}_{\mathrm{3}} \\ $$$$\:\int{tcos}\mathrm{2}{t}\:{dt} \\ $$$$={t}\frac{{sin}\mathrm{2}{t}}{\mathrm{2}}−\int\left\{\frac{{dt}}{{dt}\:}.\int{cos}\mathrm{2}{t}\:{dt}\right\}{dt} \\ $$$$={t}\frac{{sin}\mathrm{2}{t}}{\mathrm{2}}−\int\frac{{sin}\mathrm{2}{t}}{\mathrm{2}}{dt} \\ $$$$={t}\frac{{sin}\mathrm{2}{t}}{\mathrm{2}}+\frac{{cos}\mathrm{2}{t}}{\mathrm{4}} \\ $$$${so}\:{I}_{\mathrm{3}} =\mid\frac{{tsin}\mathrm{2}{t}}{\mathrm{2}}+\frac{{cos}\mathrm{2}{t}}{\mathrm{4}}\mid_{\mathrm{0}} ^{\Pi/\mathrm{4}} \\ $$

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