du-u-2-1-u-

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

\int \frac{du}{\sqrt{u^{2} - 1} - u}

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

let u = \sec θ \Rightarrow du = \sec θ \tan θ d θ

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

\int \frac{\sec θ \tan θ}{\tan θ - \sec θ} d θ =

\int \frac{\sec θ \tan θ (\tan θ + \sec θ)}{- 1} d θ

- [\int \sec θ \tan^{2} θ + \tan θ \sec^{2} θ d θ]

= - \int \sec θ (\sec^{2} θ - 1) d θ - \frac{1}{2} \tan^{2} θ

= - \int \sec θ d (\tan θ) + \ln ∣ \sec θ + \tan θ ∣ - \frac{1}{2} \tan^{2} θ

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

it easy to solving

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

I = \int \frac{d u}{\sqrt{u^{2} - 1} - u} w e u s e t h e c h a n g e m e n t u = c h (t) \Rightarrow

I = \int \frac{s h (t) d t}{s h (t) - c h (t)} = \int \frac{\frac{e^{t} - e^{- t}}{2}}{\frac{e^{t} - e^{- t}}{2} - \frac{e^{t} + e^{- t}}{2}} d t

= \int \frac{e^{t} - e^{- t}}{e^{t} - e^{- t} - e^{t} - e^{- t}} d t = \int \frac{e^{t} - e^{- t}}{- 2 e^{- t}} d t = \int \frac{e^{2 t} - 1}{- 2} d t

= - \frac{1}{2} \int (e^{2 t} - 1) d t = - \frac{1}{4} e^{2 t} + \frac{t}{2} + C

t = a r g c h (u) = l n (u + \sqrt{u^{2} - 1}) \Rightarrow e^{2 t} = {(u + \sqrt{u^{2} - 1})}^{2} \Rightarrow

I = - \frac{1}{4} {(u + \sqrt{u^{2} - 1})}^{2} + \frac{1}{2} l n (u + \sqrt{u^{2} - 1}) + C