Question Number 48667 by maxmathsup by imad last updated on 26/Nov/18
$${find}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{arctan}\left({x}\right)}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:{dx}\:. \\ $$
Commented by Abdo msup. last updated on 02/Dec/18
![changement x=tant give I =∫_0 ^(π/4) (t/(√(1+tan^2 t))) (1+tan^2 t)dt =∫_0 ^(π/4) t (dt/(cost)) =∫_0 ^(π/4) (t/(cost)) dt =_(tan((t/2))=u) ∫_0 ^((√2)−1) ((2arctanu)/((1−u^2 )/(1+u^2 ))) ((2du)/(1+u^2 )) =4 ∫_0 ^((√2)−1) ((arctan(u))/(1−u^2 )) du =4 ∫_0 ^((√2)−1) arctan(u)(Σ_(n=0) ^∞ u^(2n) )du =4 Σ_(n=0) ^∞ ∫_0 ^((√2)−1) u^(2n) arctan u du =4 Σ_(n=0) ^∞ A_n by parts A_n =∫_0 ^((√2)−1) u^(2n) arctan(u)du =[(1/(2n+1))u^(2n+1) arctanu]_0 ^((√2) −1) +∫_0 ^((√2)−1) (1/(2n+1)) u^(2n+1) (du/(1+u^2 )) =(1/(2n+1))((√2)−1)^(2n+1) arctan((√2)−1) +(1/(2n+1))∫_0 ^((√2)−1) (u^(2n+1) /(1+u^2 ))du =((π((√2)−1)^(2n+1) )/(8(2n+1))) +(1/(2n+1)) ∫_0 ^((√2)−1) (u^(2n+1) /(1+u^2 )) du but ∫_0 ^((√2)−1) (u^(2n+1) /(1+u^2 )) du =_(tanθ=u) ∫_0 ^(π/8) ((tan^(2n+1) θ)/(1+tan^2 θ)) (1+tan^2 θ)dθ = ∫_0 ^(π/8) tan^(2n+1) θ dθ ....be continued....](Q49110.png)
$${changement}\:{x}={tant}\:{give} \\ $$$${I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{t}}{\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} {t}}}\:\left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\right){dt}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{t}\:\frac{{dt}}{{cost}} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\frac{{t}}{{cost}}\:{dt}\:=_{{tan}\left(\frac{{t}}{\mathrm{2}}\right)={u}} \:\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\:\:\frac{\mathrm{2}{arctanu}}{\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }}\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\mathrm{4}\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\:\frac{{arctan}\left({u}\right)}{\mathrm{1}−{u}^{\mathrm{2}} }\:{du}\:=\mathrm{4}\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} {arctan}\left({u}\right)\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:{u}^{\mathrm{2}{n}} \right){du} \\ $$$$=\mathrm{4}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:{u}^{\mathrm{2}{n}} \:{arctan}\:{u}\:{du}\:=\mathrm{4}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{A}_{{n}} \:\:\:{by}\:{parts} \\ $$$${A}_{{n}} =\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:{u}^{\mathrm{2}{n}} \:{arctan}\left({u}\right){du}\:=\left[\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}{u}^{\mathrm{2}{n}+\mathrm{1}} \:{arctanu}\right]_{\mathrm{0}} ^{\sqrt{\mathrm{2}}\:−\mathrm{1}} \\ $$$$+\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\:{u}^{\mathrm{2}{n}+\mathrm{1}} \:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}{n}+\mathrm{1}} \:{arctan}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\:+\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\frac{{u}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{1}+{u}^{\mathrm{2}} }{du} \\ $$$$=\frac{\pi\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{8}\left(\mathrm{2}{n}+\mathrm{1}\right)}\:+\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\frac{{u}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{1}+{u}^{\mathrm{2}} }\:{du}\:{but} \\ $$$$\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\:\frac{{u}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{1}+{u}^{\mathrm{2}} }\:{du}\:\:=_{{tan}\theta={u}} \:\:\:\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{8}}} \:\:\:\frac{{tan}^{\mathrm{2}{n}+\mathrm{1}} \theta}{\mathrm{1}+{tan}^{\mathrm{2}} \theta}\:\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right){d}\theta \\ $$$$=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{8}}} \:{tan}^{\mathrm{2}{n}+\mathrm{1}} \theta\:{d}\theta\:\:....{be}\:{continued}.... \\ $$
Answered by Abdulhafeez Abu qatada last updated on 26/Nov/18
