find-2-5-dt-t-t-2-1-

Question Number 31503 by abdo imad last updated on 09/Mar/18

f i n d \int_{2}^{\sqrt{5}} \frac{d t}{t \sqrt{t^{2} - 1}} .

Commented by abdo imad last updated on 16/Mar/18

c h . t = c h (x) g i v e I = \int_{a r g c h (2)}^{a r g c h (\sqrt{5})} \frac{s h x d x}{c h (x) s h (x)}

= \int_{a r g c h (2)}^{a r g c h (\sqrt{5})} \frac{d x}{\frac{e^{x} + e^{- x}}{2}} = 2 \int_{a r g c h 2)}^{a r g c h (\sqrt{5})} \frac{d x}{e^{x} + e^{- x}}

l e t u s e t h e c h . e^{x} = t \Rightarrow I = 2 \int_{e^{a r g c h (2)}}^{e^{a r g c h (\sqrt{5})}} \frac{d t}{t (t + \frac{1}{t})}

= 2 \int_{e^{a r g c h (2)}}^{e^{a r g c h (\sqrt{5})}} \frac{d t}{t^{2} + 1} = 2 {[a r c t a n (t)]}_{e^{a r g c h (2)}}^{e^{a r g c h (\sqrt{5})}} b u t w e k n o w

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

a r g c h (2) = l n (2 + \sqrt{3}) a n d a r g c h (\sqrt{5}) = l n (\sqrt{5} + 2)

I = 2 {[a r c t a n (t)]}_{2 + \sqrt{3}}^{2 + \sqrt{5}} = 2 (a r c t a n (2 + \sqrt{5}) - a r c t a n (2 + \sqrt{3}))