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Question Number 76143 by Master last updated on 24/Dec/19

Commented by mathmax by abdo last updated on 24/Dec/19
$let I =∫_0 ^1 (√(4+x^2 ))dx changement x=2sh(t) give I =∫_0 ^(argsh((1/2))) 2ch(t)2ch(t)dt =4 ∫_0 ^(ln((1/2)+(√(1+(1/4))))) ch^2 (t)dt =4 ∫_0 ^(ln((1/2)+((√5)/2))) ((ch(2t)−1)/2)dt =2 ∫_0 ^(ln(((1+(√5))/2))) (ch(2t)−1)dt =[sh(2t)]_0 ^(ln(((1+(√5))/2))) −2ln(((1+(√5))/2)) =[((e^(2t) −e^(−2t) )/2)]_0 ^(ln(((1+(√5))/2))) −2ln(((1+(√5))/2)) =(1/2){ (((1+(√5))/2))^2 −(((1+(√5))/2))^(−2) }−2ln(((1+(√5))/2))$
$l e t I = \int_{0}^{1} \sqrt{4 + x^{2}} d x c h a n g e m e n t x = 2 s h (t) g i v e$ $I = \int_{0}^{a r g s h (\frac{1}{2})} 2 c h (t) 2 c h (t) d t = 4 \int_{0}^{l n (\frac{1}{2} + \sqrt{1 + \frac{1}{4}})} {c h}^{2} (t) d t$ $= 4 \int_{0}^{l n (\frac{1}{2} + \frac{\sqrt{5}}{2})} \frac{c h (2 t) - 1}{2} d t = 2 \int_{0}^{l n (\frac{1 + \sqrt{5}}{2})} (c h (2 t) - 1) d t$ $= {[s h (2 t)]}_{0}^{l n (\frac{1 + \sqrt{5}}{2})} - 2 l n (\frac{1 + \sqrt{5}}{2})$ $= {[\frac{e^{2 t} - e^{- 2 t}}{2}]}_{0}^{l n (\frac{1 + \sqrt{5}}{2})} - 2 l n (\frac{1 + \sqrt{5}}{2})$ $= \frac{1}{2} {{(\frac{1 + \sqrt{5}}{2})}^{2} - {(\frac{1 + \sqrt{5}}{2})}^{- 2}} - 2 l n (\frac{1 + \sqrt{5}}{2})$
	Answered by john santuy last updated on 24/Dec/19

	$b y l e t t i n g x = 2 t a n t \to d x = 2 {s e c}^{2} t d t .$ $t h e n \underset{0}{\int^{\frac{π}{4}}} 2 {s e c}^{2} t \times 2 s e c t d t = \underset{0}{\int^{\frac{π}{4}}} 4 s e c t d (t a n t)$ $n o w c a n s o l v e w i t h i n t e g r a t i o n b y p a r t$
	Answered by benjo last updated on 24/Dec/19

	Commented by Master last updated on 24/Dec/19

	$thanks$
	Commented by benjo last updated on 24/Dec/19

	$okay sir$
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