find g(a) =∫(x+a)(√(x^2 −a^2 ))dx


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Question Number 62209 by maxmathsup by imad last updated on 17/Jun/19

$f i n d g (a) = \int (x + a) \sqrt{x^{2} - a^{2}} d x$
Commented by maxmathsup by imad last updated on 18/Jun/19
$changement x=ach(t) give g(a) =∫ (acht +a)(√(a^2 (ch^2 t−1)))ash(t)dt =a^2 ∣a∣ ∫ (1+ch(t)sh^2 (t) dt =a^2 ∣a∣ ∫2ch^2 t sh^2 t dt =2a^2 ∣a∣ ∫ ((1/2)sh(2t))^2 dt =((a^2 ∣a∣)/2) ∫ ((ch(4t)−1)/2)dt =((a^2 ∣a∣)/4) ∫ (ch(4t)−1)dt =((a^2 ∣a∣)/(16)) sh(4t)−((a^2 ∣a∣)/4) t +c but we have ch(t)=(x/a) ⇒t =argch((x/a))=ln((x/a) +(√((x^2 /a^2 )−1))) ⇒ 4t =ln{ ((x/a) +(√((x^2 /a^2 )−1)))^4 } and sh(4t) =((e^(4t) −e^(−4t) )/2) =(1/2){ ((x/a)+(√((x^2 /a^2 )−1)))^4 −((x/a)+(√((x^2 /a^2 )−1)))^(−4) } ⇒ A =((a^2 ∣a∣)/(32)){ ((x/a) +(√((x^2 /a^2 )−1)))^4 −((x/a)+(√(x^2 /a^2 ))−1)^(−4) }−((a^2 ∣a∣)/4)ln((x/a)+(√((x^2 /a^2 )−1))) +C$
$c h a n g e m e n t x = a c h (t) g i v e g (a) = \int (a c h t + a) \sqrt{a^{2} ({c h}^{2} t - 1)} a s h (t) d t$ $= a^{2} ∣ a ∣ \int (1 + c h (t) {s h}^{2} (t) d t$ $= a^{2} ∣ a ∣ \int 2 {c h}^{2} t {s h}^{2} t d t = 2 a^{2} ∣ a ∣ \int {(\frac{1}{2} s h (2 t))}^{2} d t$ $= \frac{a^{2} ∣ a ∣}{2} \int \frac{c h (4 t) - 1}{2} d t = \frac{a^{2} ∣ a ∣}{4} \int (c h (4 t) - 1) d t$ $= \frac{a^{2} ∣ a ∣}{16} s h (4 t) - \frac{a^{2} ∣ a ∣}{4} t + c b u t w e h a v e$ $c h (t) = \frac{x}{a} \Rightarrow t = a r g c h (\frac{x}{a}) = l n (\frac{x}{a} + \sqrt{\frac{x^{2}}{a^{2}} - 1}) \Rightarrow$ $4 t = l n {{(\frac{x}{a} + \sqrt{\frac{x^{2}}{a^{2}} - 1})}^{4}} a n d$ $s h (4 t) = \frac{e^{4 t} - e^{- 4 t}}{2} = \frac{1}{2} {{(\frac{x}{a} + \sqrt{\frac{x^{2}}{a^{2}} - 1})}^{4} - {(\frac{x}{a} + \sqrt{\frac{x^{2}}{a^{2}} - 1})}^{- 4}} \Rightarrow$ $A = \frac{a^{2} ∣ a ∣}{32} {{(\frac{x}{a} + \sqrt{\frac{x^{2}}{a^{2}} - 1})}^{4} - {(\frac{x}{a} + \sqrt{\frac{x^{2}}{a^{2}}} - 1)}^{- 4}} - \frac{a^{2} ∣ a ∣}{4} l n (\frac{x}{a} + \sqrt{\frac{x^{2}}{a^{2}} - 1}) + C$
	Answered by tanmay last updated on 17/Jun/19

	$\int x \sqrt{x^{2} - a^{2}} d x + a \int \sqrt{x^{2} - a^{2}} d x$ $\frac{1}{2} \int {(x^{2} - a^{2})}^{\frac{1}{2}} d (x^{2} - a^{2}) + a \int \sqrt{x^{2} - a^{2}} d x$ $\frac{1}{2} \times \frac{{(x^{2} - a^{2})}^{\frac{3}{2}}}{\frac{3}{2}} + a [\frac{x \sqrt{x^{2} - a^{2}}}{2} - \frac{a^{2}}{2} l n (x + \sqrt{x^{2} - a^{2}})] + c$
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