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Question Number 179510 by Acem last updated on 30/Oct/22
Evaluate ∫e^(4x) (√((1/e^(−2x) )+1)) dx
$${Evaluate}\:\int{e}^{\mathrm{4}{x}} \:\sqrt{\frac{\mathrm{1}}{{e}^{−\mathrm{2}{x}} }+\mathrm{1}}\:{dx} \\ $$
Commented by cortano1 last updated on 30/Oct/22
(√(e^(2x) +1)) = λ ⇒e^(2x) =λ^2 −1 e^(2x) dx = λ dλ ⇒I = ∫ e^(2x) (√(e^(2x) +1)) (e^(2x) dx) ⇒I= ∫ (λ^2 −1)(λ)(λ dλ) ⇒I=∫ (λ^4 −λ^2 )dλ = (1/5)λ^5 −(1/3)λ^3 +c ⇒I=(1/(15))λ^3 (3λ^2 −5)+c ⇒I=(1/(15))(√((e^(2x) +1)^3 )) (3e^(2x) −2)+c
$$\sqrt{\mathrm{e}^{\mathrm{2x}} +\mathrm{1}}\:=\:\lambda\:\Rightarrow\mathrm{e}^{\mathrm{2x}} =\lambda^{\mathrm{2}} −\mathrm{1} \\ $$$$\:\mathrm{e}^{\mathrm{2x}} \:\mathrm{dx}\:=\:\lambda\:\mathrm{d}\lambda \\ $$$$\:\Rightarrow\mathrm{I}\:=\:\int\:\mathrm{e}^{\mathrm{2x}} \:\sqrt{\mathrm{e}^{\mathrm{2x}} +\mathrm{1}}\:\left(\mathrm{e}^{\mathrm{2x}} \:\mathrm{dx}\right) \\ $$$$\Rightarrow\mathrm{I}=\:\int\:\left(\lambda^{\mathrm{2}} −\mathrm{1}\right)\left(\lambda\right)\left(\lambda\:\mathrm{d}\lambda\right) \\ $$$$\Rightarrow\mathrm{I}=\int\:\left(\lambda^{\mathrm{4}} −\lambda^{\mathrm{2}} \right)\mathrm{d}\lambda\:=\:\frac{\mathrm{1}}{\mathrm{5}}\lambda^{\mathrm{5}} −\frac{\mathrm{1}}{\mathrm{3}}\lambda^{\mathrm{3}} +\mathrm{c} \\ $$$$\Rightarrow\mathrm{I}=\frac{\mathrm{1}}{\mathrm{15}}\lambda^{\mathrm{3}} \left(\mathrm{3}\lambda^{\mathrm{2}} −\mathrm{5}\right)+\mathrm{c} \\ $$$$\Rightarrow\mathrm{I}=\frac{\mathrm{1}}{\mathrm{15}}\sqrt{\left(\mathrm{e}^{\mathrm{2x}} +\mathrm{1}\right)^{\mathrm{3}} }\:\left(\mathrm{3e}^{\mathrm{2x}} −\mathrm{2}\right)+\mathrm{c}\: \\ $$
Commented by Acem last updated on 30/Oct/22
Great Sir!
$${Great}\:{Sir}! \\ $$
Answered by Acem last updated on 30/Oct/22
I= ∫e^(4x) (√(e^(2x) +1)) dx e^x = tan θ , e^x dx= sec^2 θ dθ (√(e^(2x) +1)) = ∣sec θ∣= sec θ^( Bec. Indefinite integral) I= ∫tan^3 θ sec^3 θ dθ I= ∫(sec^2 θ−1) sec^2 θ d(sec θ) = (1/5) sec^5 θ −(1/3) sec^3 θ + c = (1/(15)) sec^3 θ (3 sec^2 θ −5) +c Since tan θ= e^x ⇒ sec θ= (√(e^(2x) +1)) I= (1/(15)) (√((e^(2x) +1)^3 )) (3 e^(2x) −2) + c
$$ \\ $$$$\:{I}=\:\int{e}^{\mathrm{4}{x}} \:\sqrt{{e}^{\mathrm{2}{x}} +\mathrm{1}}\:{dx} \\ $$$$\:{e}^{{x}} =\:\mathrm{tan}\:\theta\:,\:{e}^{{x}} \:{dx}=\:\mathrm{sec}^{\mathrm{2}} \:\theta\:{d}\theta \\ $$$$\:\sqrt{{e}^{\mathrm{2}{x}} +\mathrm{1}}\:=\:\mid\mathrm{sec}\:\theta\mid=\:\mathrm{sec}\:\theta^{\:{Bec}.\:{Indefinite}\:{integral}} \\ $$$$\:{I}=\:\int\mathrm{tan}^{\mathrm{3}} \:\theta\:\mathrm{sec}^{\mathrm{3}} \:\theta\:{d}\theta \\ $$$$\:{I}=\:\int\left(\mathrm{sec}^{\mathrm{2}} \:\theta−\mathrm{1}\right)\:\mathrm{sec}^{\mathrm{2}} \:\theta\:{d}\left(\mathrm{sec}\:\theta\right) \\ $$$$\:\:\:=\:\frac{\mathrm{1}}{\mathrm{5}}\:\mathrm{sec}^{\mathrm{5}} \:\theta\:−\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{sec}^{\mathrm{3}} \:\theta\:+\:{c} \\ $$$$\:\:\:=\:\frac{\mathrm{1}}{\mathrm{15}}\:\mathrm{sec}^{\mathrm{3}} \:\theta\:\left(\mathrm{3}\:\mathrm{sec}^{\mathrm{2}} \:\theta\:−\mathrm{5}\right)\:+{c} \\ $$$$\:{Since}\:\mathrm{tan}\:\theta=\:{e}^{{x}} \:\Rightarrow\:\mathrm{sec}\:\theta=\:\sqrt{{e}^{\mathrm{2}{x}} +\mathrm{1}} \\ $$$$\:{I}=\:\frac{\mathrm{1}}{\mathrm{15}}\:\sqrt{\left({e}^{\mathrm{2}{x}} +\mathrm{1}\right)^{\mathrm{3}} }\:\left(\mathrm{3}\:{e}^{\mathrm{2}{x}} \:−\mathrm{2}\right)\:+\:{c} \\ $$$$ \\ $$

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