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Question Number 27797 by abdo imad last updated on 14/Jan/18
find ∫ (√(2+tan^2 t)) dt.
$${find}\:\:\:\int\:\sqrt{\mathrm{2}+{tan}^{\mathrm{2}} {t}}\:\:{dt}. \\ $$
Commented by abdo imad last updated on 19/Jan/18
let put I= ∫(√(2+tan^2 t))dt I= ∫(√(1+tan^2 +1))dt =∫ (√( (1/(cos^2 t)) +1))dt = ∫ ((√(1+cos^2 t))/(cost))dt the ch. (√(1+cos^2 t)) =x give 1+cos^2 t =x^2 ⇒ cos^2 t= x^2 −1 ⇒cost=(√(x^2 −1)) ⇒t= arcos((√(x^2 −1)))⇒ dt= ((−x)/( (√(x^2 −1))(√(1−(x^2 −1)))))dx = ((−x)/( (√(x^2 −1))(√(2−x^2 )))) I= ∫ (x/( (√(x^2 −1)))) ((−xdx)/( (√(x^2 −1))(√(2−x^2 ))))= −∫ (x^2 /((x^2 −1)(√(2−x^2 ))))dx = − ∫ ((x^2 −1 +1)/((x^2 −1)(√(2−x^2 ))))dx =− ∫ (dx/( (√(2−x^2 )))) −∫ (dx/((x^2 −1)(√(2−x^2 )))) but ∫ (dx/( (√(2−x^2 )))) = ∫ (((√(2 ))dt)/( (√2)(√(1−t^2 )))) =arcsin ((x/( (√2)))) +k and −∫ (dx/((x^2 −1)(√(2−x^2 ))))= ∫ (dx/((1−x^2 )(√(2−x^2 )))) = ∫ (dx/((2−x^2 −1)(√(2−x^2 )))) the ch. (√(2−x^2 ))=t give 2−x^2 =t^2 ⇒ x^2 =2−t^2 ⇒ x =(√(2−t^2 )) ⇒dx= ((−t)/( (√(2−t^2 ))))dt = ∫ ((−t)/( (√(2−t^2 ))(t^2 −1)t))dt = ∫ (dt/((t^2 −1)(√(2−x^2 )))).......
$${let}\:{put}\:\:{I}=\:\int\sqrt{\mathrm{2}+{tan}^{\mathrm{2}} {t}}{dt}\:\:\:\: \\ $$$${I}=\:\int\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} +\mathrm{1}}{dt}\:\:=\int\:\sqrt{\:\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {t}}\:+\mathrm{1}}{dt} \\ $$$$=\:\:\int\:\:\:\frac{\sqrt{\mathrm{1}+{cos}^{\mathrm{2}} {t}}}{{cost}}{dt}\:\:\:{the}\:{ch}.\:\sqrt{\mathrm{1}+{cos}^{\mathrm{2}} {t}}\:\:={x}\:{give} \\ $$$$\mathrm{1}+{cos}^{\mathrm{2}} {t}\:={x}^{\mathrm{2}} \:\Rightarrow\:{cos}^{\mathrm{2}} {t}=\:{x}^{\mathrm{2}} −\mathrm{1}\:\:\Rightarrow{cost}=\sqrt{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\Rightarrow{t}=\:{arcos}\left(\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)\Rightarrow\:\:{dt}=\:\:\:\frac{−{x}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\sqrt{\mathrm{1}−\left({x}^{\mathrm{2}} −\mathrm{1}\right)}}{dx} \\ $$$$=\:\:\frac{−{x}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\sqrt{\mathrm{2}−{x}^{\mathrm{2}} }} \\ $$$${I}=\:\:\int\:\:\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\:\:\frac{−{xdx}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\sqrt{\mathrm{2}−{x}^{\mathrm{2}} }}=\:−\int\:\frac{{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} −\mathrm{1}\right)\sqrt{\mathrm{2}−{x}^{\mathrm{2}} }}{dx} \\ $$$$=\:−\:\int\:\:\frac{{x}^{\mathrm{2}} −\mathrm{1}\:+\mathrm{1}}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)\sqrt{\mathrm{2}−{x}^{\mathrm{2}} }}{dx}\:\:=−\:\int\:\:\frac{{dx}}{\:\sqrt{\mathrm{2}−{x}^{\mathrm{2}} }}\:−\int\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)\sqrt{\mathrm{2}−{x}^{\mathrm{2}} }} \\ $$$$\:{but}\:\:\:\int\:\:\frac{{dx}}{\:\sqrt{\mathrm{2}−{x}^{\mathrm{2}} }}\:=\:\:\int\:\:\:\:\frac{\sqrt{\mathrm{2}\:}{dt}}{\:\sqrt{\mathrm{2}}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:={arcsin}\:\left(\frac{{x}}{\:\sqrt{\mathrm{2}}}\right)\:+{k} \\ $$$${and}\:−\int\:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)\sqrt{\mathrm{2}−{x}^{\mathrm{2}} }}=\:\int\:\:\:\frac{{dx}}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\sqrt{\mathrm{2}−{x}^{\mathrm{2}} }} \\ $$$$=\:\int\:\:\:\:\:\frac{{dx}}{\left(\mathrm{2}−{x}^{\mathrm{2}} −\mathrm{1}\right)\sqrt{\mathrm{2}−{x}^{\mathrm{2}} }}\:{the}\:{ch}.\:\:\:\sqrt{\mathrm{2}−{x}^{\mathrm{2}} }={t}\:{give} \\ $$$$\mathrm{2}−{x}^{\mathrm{2}} \:={t}^{\mathrm{2}} \:\Rightarrow\:{x}^{\mathrm{2}} =\mathrm{2}−{t}^{\mathrm{2}} \:\Rightarrow\:{x}\:=\sqrt{\mathrm{2}−{t}^{\mathrm{2}} \:}\:\:\Rightarrow{dx}=\:\frac{−{t}}{\:\sqrt{\mathrm{2}−{t}^{\mathrm{2}} }}{dt} \\ $$$$=\:\int\:\:\:\:\:\frac{−{t}}{\:\sqrt{\mathrm{2}−{t}^{\mathrm{2}} }\left({t}^{\mathrm{2}} −\mathrm{1}\right){t}}{dt}\:=\:\:\int\:\:\:\:\frac{{dt}}{\left({t}^{\mathrm{2}} −\mathrm{1}\right)\sqrt{\mathrm{2}−{x}^{\mathrm{2}} }}……. \\ $$

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