find ∫_(1/2) ^1 (dx/((√(4x^2 −1)) +(√(4x^2 +1))))


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Question Number 40146 by maxmathsup by imad last updated on 16/Jul/18

$f i n d \int_{\frac{1}{2}}^{1} \frac{d x}{\sqrt{4 x^{2} - 1} + \sqrt{4 x^{2} + 1}}$
Commented by maxmathsup by imad last updated on 19/Jul/18
$let I = ∫_(1/2) ^1 (dx/((√(4x^2 −1)) −(√(4x^2 +1)))) changement 2x=t give I = ∫_1 ^2 (1/((√(t^2 −1)) +(√(t^2 +1)))) (dt/2) ⇒ 2I = ∫_1 ^2 (((√(t^2 +1)) −(√(t^2 −1)))/2)dt ⇒4I = ∫_1 ^2 (√(t^2 +1))dt−∫_1 ^2 (√(t^2 −1))dt but ∫_1 ^2 (√(t^2 +1))dt =_(t=sh(x)) ∫_(argsh(1)) ^(argsh(2)) (√(1+sh^2 x))chxdx= ∫_(ln(1+(√2))) ^(ln(2+(√5))) ch^2 xdx =(1/2) ∫_(ln(1+(√2))) ^(ln(2+(√5))) (1+ch(2x))dx =(1/2)(ln(2+(√5))−ln(1+(√2)) +(1/4) [sh(2x)]_(ln(1+(√2))) ^(ln(2+(√5))) =(1/2){ln(2+(√5)) −ln(1+(√2))} +(1/4)[((e^(2x) −e^(−2x) )/2)]_(ln(1+(√2))) ^(ln(2+(√5))) =(1/2){ln(2+(√5))−ln(1+(√2))} +(1/8){(2+(√5))^2 −(2+(√5))^(−2) −(1+(√2))^2 +(1+(√2))^(−2) } also changement t =ch(x) give ∫_1 ^2 (√(t^2 −1))dt = ∫_(argch(1)) ^(argch(2)) sh(x)sh(x)dx = ∫_0 ^(ln(2+(√3))) ((ch(2x)−1)/2)dx=−(1/2)ln(2+(√3)) +(1/4)[sh(2x)^ ]_0 ^(ln(2+(√3))) =−(1/2)ln(2+(√3)) +(1/8){ (2+(√3))^2 −(2+(√3))^(−2) } .$
$l e t I = \int_{\frac{1}{2}}^{1} \frac{d x}{\sqrt{4 x^{2} - 1} - \sqrt{4 x^{2} + 1}} c h a n g e m e n t 2 x = t g i v e$ $I = \int_{1}^{2} \frac{1}{\sqrt{t^{2} - 1} + \sqrt{t^{2} + 1}} \frac{d t}{2} \Rightarrow$ $2 I = \int_{1}^{2} \frac{\sqrt{t^{2} + 1} - \sqrt{t^{2} - 1}}{2} d t \Rightarrow 4 I = \int_{1}^{2} \sqrt{t^{2} + 1} d t - \int_{1}^{2} \sqrt{t^{2} - 1} d t b u t$ $\int_{1}^{2} \sqrt{t^{2} + 1} d t =_{t = s h (x)} \int_{a r g s h (1)}^{a r g s h (2)} \sqrt{1 + {s h}^{2} x} c h x d x = \int_{l n (1 + \sqrt{2})}^{l n (2 + \sqrt{5})} {c h}^{2} x d x$ $= \frac{1}{2} \int_{l n (1 + \sqrt{2})}^{l n (2 + \sqrt{5})} (1 + c h (2 x)) d x$ $= \frac{1}{2} (l n (2 + \sqrt{5}) - l n (1 + \sqrt{2}) + \frac{1}{4} {[s h (2 x)]}_{l n (1 + \sqrt{2})}^{l n (2 + \sqrt{5})}$ $= \frac{1}{2} {l n (2 + \sqrt{5}) - l n (1 + \sqrt{2})} + \frac{1}{4} {[\frac{e^{2 x} - e^{- 2 x}}{2}]}_{l n (1 + \sqrt{2})}^{l n (2 + \sqrt{5})}$ $= \frac{1}{2} {l n (2 + \sqrt{5}) - l n (1 + \sqrt{2})} + \frac{1}{8} {{(2 + \sqrt{5})}^{2} - {(2 + \sqrt{5})}^{- 2} - {(1 + \sqrt{2})}^{2} + {(1 + \sqrt{2})}^{- 2}}$ $a l s o c h a n g e m e n t t = c h (x) g i v e$ $\int_{1}^{2} \sqrt{t^{2} - 1} d t = \int_{a r g c h (1)}^{a r g c h (2)} s h (x) s h (x) d x$ $Missing \left or extra \right$ $= - \frac{1}{2} l n (2 + \sqrt{3}) + \frac{1}{8} {{(2 + \sqrt{3})}^{2} - {(2 + \sqrt{3})}^{- 2}} .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 17/Jul/18

	$\int_{\frac{1}{2}}^{1} \frac{\sqrt{4 x^{2} + 1} - \sqrt{4 x^{2} - 1}}{2} d x$ $\frac{1}{2} \int_{\frac{1}{2}}^{1} 2 \sqrt{x^{2} + {(\frac{1}{2})}^{2}} d x - \frac{1}{2} \int_{\frac{1}{2}}^{1} 2 \sqrt{x^{2} - {(\frac{1}{2})}^{2}} d x$ $n o w j s e f o r m u l a$ $\int \sqrt{x^{2} + a^{2}} d x a n d \int \sqrt{x^{2} - a^{2}} d x$
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