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Question Number 83898 by sahnaz last updated on 07/Mar/20

∫(du/((√(u^2 −1 ))−u))

$$\int\frac{\mathrm{du}}{\sqrt{\mathrm{u}^{\mathrm{2}} −\mathrm{1}\:}−\mathrm{u}} \\ $$

Commented by john santu last updated on 07/Mar/20

let u = sec θ ⇒ du = sec θ tan θ dθ

$$\mathrm{let}\:{u}\:=\:\mathrm{sec}\:\theta\:\Rightarrow\:\mathrm{du}\:=\:\mathrm{sec}\:\theta\:\mathrm{tan}\:\theta\:\mathrm{d}\theta \\ $$

Commented by john santu last updated on 07/Mar/20

∫ ((sec θ tan θ)/(tan θ − sec θ)) dθ = ∫ ((sec θ tan θ (tan θ+sec θ))/(−1)) dθ −[∫sec θ tan^2 θ + tan θ sec^2 θ dθ] =−∫ sec θ (sec^2 θ−1)dθ−(1/2)tan^2 θ = −∫sec θ d(tan θ)+ln ∣sec θ+tan θ∣−(1/2)tan^2 θ

$$\int\:\frac{\mathrm{sec}\:\theta\:\mathrm{tan}\:\theta}{\mathrm{tan}\:\theta\:−\:\mathrm{sec}\:\theta}\:\mathrm{d}\theta\:=\: \\ $$$$\int\:\frac{\mathrm{sec}\:\theta\:\mathrm{tan}\:\theta\:\left(\mathrm{tan}\:\theta+\mathrm{sec}\:\theta\right)}{−\mathrm{1}}\:\mathrm{d}\theta \\ $$$$−\left[\int\mathrm{sec}\:\theta\:\mathrm{tan}\:^{\mathrm{2}} \theta\:+\:\mathrm{tan}\:\theta\:\mathrm{sec}\:^{\mathrm{2}} \theta\:\mathrm{d}\theta\right] \\ $$$$=−\int\:\mathrm{sec}\:\theta\:\left(\mathrm{sec}\:^{\mathrm{2}} \theta−\mathrm{1}\right)\mathrm{d}\theta−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}\:^{\mathrm{2}} \theta\: \\ $$$$=\:−\int\mathrm{sec}\:\theta\:\mathrm{d}\left(\mathrm{tan}\:\theta\right)+\mathrm{ln}\:\mid\mathrm{sec}\:\theta+\mathrm{tan}\:\theta\mid−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}\:^{\mathrm{2}} \theta \\ $$$$ \\ $$

Commented by john santu last updated on 07/Mar/20

it easy to solving

$$\mathrm{it}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{solving} \\ $$

Commented by mathmax by abdo last updated on 07/Mar/20

I=∫ (du/((√(u^2 −1))−u)) we use the changement u=ch(t) ⇒ I =∫ ((sh(t)dt)/(sh(t)−ch(t))) =∫ (((e^t −e^(−t) )/2)/(((e^t −e^(−t) )/2)−((e^t +e^(−t) )/2)))dt =∫ ((e^t −e^(−t) )/(e^t −e^(−t) −e^t −e^(−t) )) dt =∫ ((e^t −e^(−t) )/(−2e^(−t) ))dt =∫((e^(2t) −1)/(−2))dt =−(1/2)∫ (e^(2t) −1)dt =−(1/4)e^(2t) +(t/2) +C t=argch(u) =ln(u+(√(u^2 −1))) ⇒e^(2t) =(u+(√(u^2 −1)))^2 ⇒ I =−(1/4)(u+(√(u^2 −1)))^2 +(1/2)ln(u+(√(u^2 −1)))+C

$${I}=\int\:\frac{{du}}{\sqrt{{u}^{\mathrm{2}} −\mathrm{1}}−{u}}\:\:\:{we}\:{use}\:{the}\:{changement}\:{u}={ch}\left({t}\right)\:\Rightarrow \\ $$$${I}\:=\int\:\:\frac{{sh}\left({t}\right){dt}}{{sh}\left({t}\right)−{ch}\left({t}\right)}\:=\int\:\frac{\frac{{e}^{{t}} −{e}^{−{t}} }{\mathrm{2}}}{\frac{{e}^{{t}} −{e}^{−{t}} }{\mathrm{2}}−\frac{{e}^{{t}} \:+{e}^{−{t}} }{\mathrm{2}}}{dt} \\ $$$$=\int\:\:\:\frac{{e}^{{t}} −{e}^{−{t}} }{{e}^{{t}} −{e}^{−{t}} −{e}^{{t}} −{e}^{−{t}} }\:{dt}\:=\int\:\:\frac{{e}^{{t}} \:−{e}^{−{t}} }{−\mathrm{2}{e}^{−{t}} }{dt}\:\:=\int\frac{{e}^{\mathrm{2}{t}} −\mathrm{1}}{−\mathrm{2}}{dt} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\int\:\:\left({e}^{\mathrm{2}{t}} −\mathrm{1}\right){dt}\:=−\frac{\mathrm{1}}{\mathrm{4}}{e}^{\mathrm{2}{t}} \:+\frac{{t}}{\mathrm{2}}\:+{C} \\ $$$${t}={argch}\left({u}\right)\:={ln}\left({u}+\sqrt{{u}^{\mathrm{2}} −\mathrm{1}}\right)\:\Rightarrow{e}^{\mathrm{2}{t}} \:=\left({u}+\sqrt{{u}^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{2}} \:\Rightarrow \\ $$$${I}\:=−\frac{\mathrm{1}}{\mathrm{4}}\left({u}+\sqrt{{u}^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({u}+\sqrt{{u}^{\mathrm{2}} −\mathrm{1}}\right)+{C} \\ $$

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