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Question Number 40830 by math khazana by abdo last updated on 28/Jul/18
find ∫ (√(2+tan^2 t))dt
$${find}\:\int\:\sqrt{\mathrm{2}+{tan}^{\mathrm{2}} {t}}{dt} \\ $$
Commented by maxmathsup by imad last updated on 31/Jul/18
let I = ∫ (√(2+tan^2 t))dt changement tant =(√2) sh(x)⇒t =arctan((√2)shx) ⇒dt =(((√2)ch(x))/(1+2sh^2 x))dx I = ∫ (√(2 +2sh^2 x)) (((√2)ch(x))/(1+2sh^2 x))dx = ∫ ((2ch^2 x)/(1+2sh^2 ))dx = ∫ ((1+ch(2x))/(1+ch(2x)−1))dx = ∫ ((ch(2x) +1)/(ch(2x)))dx = x +∫ (dx/(ch(2x))) but ∫ (dx/(ch(2x))) = ∫ ((2dx)/(e^(2x) +e^(−2x) )) =_(e^x =t) ∫ ((2 )/(t^2 +t^(−2) )) (dt/t) = ∫ ((2dt)/(t^3 +(1/t))) = ∫ ((2tdt)/(t^4 +1)) =_(t^2 =u) ∫ (du/(1+u^2 )) =arctan(u) +c =arctan(t^2 )+c =actan(e^(2x) )+c but x =argsh(((tant)/( (√2)))) =ln(((tant)/( (√2))) +(√(1+((tan^2 t)/2))))) I =argsh(((tant)/( (√2)))) +arctan( {((tant)/( (√2)))+(√(1+((tan^2 t)/2)))}^2 ) +c
$${let}\:{I}\:\:=\:\int\:\sqrt{\mathrm{2}+{tan}^{\mathrm{2}} {t}}{dt}\:\:\:{changement}\:{tant}\:=\sqrt{\mathrm{2}}\:{sh}\left({x}\right)\Rightarrow{t}\:={arctan}\left(\sqrt{\mathrm{2}}{shx}\right) \\ $$$$\Rightarrow{dt}\:=\frac{\sqrt{\mathrm{2}}{ch}\left({x}\right)}{\mathrm{1}+\mathrm{2}{sh}^{\mathrm{2}} {x}}{dx} \\ $$$${I}\:=\:\int\:\sqrt{\mathrm{2}\:+\mathrm{2}{sh}^{\mathrm{2}} {x}}\:\frac{\sqrt{\mathrm{2}}{ch}\left({x}\right)}{\mathrm{1}+\mathrm{2}{sh}^{\mathrm{2}} {x}}{dx}\:\:=\:\int\:\:\frac{\mathrm{2}{ch}^{\mathrm{2}} {x}}{\mathrm{1}+\mathrm{2}{sh}^{\mathrm{2}} }{dx} \\ $$$$=\:\:\int\:\:\:\:\frac{\mathrm{1}+{ch}\left(\mathrm{2}{x}\right)}{\mathrm{1}+{ch}\left(\mathrm{2}{x}\right)−\mathrm{1}}{dx}\:=\:\int\:\:\:\:\frac{{ch}\left(\mathrm{2}{x}\right)\:+\mathrm{1}}{{ch}\left(\mathrm{2}{x}\right)}{dx}\:=\:{x}\:+\int\:\:\:\frac{{dx}}{{ch}\left(\mathrm{2}{x}\right)}\:{but} \\ $$$$\int\:\:\:\frac{{dx}}{{ch}\left(\mathrm{2}{x}\right)}\:=\:\int\:\:\:\:\frac{\mathrm{2}{dx}}{{e}^{\mathrm{2}{x}} \:+{e}^{−\mathrm{2}{x}} }\:=_{{e}^{{x}} ={t}} \:\:\:\:\int\:\:\frac{\mathrm{2}\:\:}{{t}^{\mathrm{2}} \:+{t}^{−\mathrm{2}} }\:\frac{{dt}}{{t}} \\ $$$$=\:\int\:\:\:\frac{\mathrm{2}{dt}}{{t}^{\mathrm{3}} \:+\frac{\mathrm{1}}{{t}}}\:=\:\int\:\:\:\frac{\mathrm{2}{tdt}}{{t}^{\mathrm{4}} \:+\mathrm{1}}\:\:=_{{t}^{\mathrm{2}} ={u}} \:\int\:\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:={arctan}\left({u}\right)\:+{c} \\ $$$$={arctan}\left({t}^{\mathrm{2}} \right)+{c}\:\:={actan}\left({e}^{\mathrm{2}{x}} \right)+{c}\:\:{but}\:{x}\:={argsh}\left(\frac{{tant}}{\:\sqrt{\mathrm{2}}}\right) \\ $$$$={ln}\left(\frac{{tant}}{\:\sqrt{\mathrm{2}}}\:+\sqrt{\left.\mathrm{1}+\frac{{tan}^{\mathrm{2}} {t}}{\mathrm{2}}\right)}\right) \\ $$$${I}\:\:={argsh}\left(\frac{{tant}}{\:\sqrt{\mathrm{2}}}\right)\:\:+{arctan}\left(\:\left\{\frac{{tant}}{\:\sqrt{\mathrm{2}}}+\sqrt{\mathrm{1}+\frac{{tan}^{\mathrm{2}} {t}}{\mathrm{2}}}\right\}^{\mathrm{2}} \right)\:+{c} \\ $$

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