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Question Number 82330 by Power last updated on 20/Feb/20

Commented by mathmax by abdo last updated on 20/Feb/20
$let I =∫x^2 (√(a^2 +x^2 ))dx changement x=ash(t) give I =∫ a^2 sh^2 (t)∣a∣ch(t)ach(t)dt =a^3 ∣a∣∫sh^2 (t)ch^2 (t)dt =a^3 ∣a∣ ∫( ((ch(2t)−1)/2))^2 ×(((1+ch(2t))/2))^2 dt =((a^3 ∣a∣)/(16)) ∫(ch(2t)−1)(ch(2t)+1)dt =((a^3 ∣a∣)/(16)) ∫(ch^2 (2t)−1)dt =((a^3 ∣a∣)/(16)) ∫(((1+ch(4t))/2)−1)dt =((a^3 ∣a∣)/(32))∫ (ch(4t)−1)dt =((a^3 ∣a∣)/(32×4))sh(4t)−((a^3 ∣a∣)/(32))t +C sh(4t) =((e^(4t) −e^(−4t) )/2) and t=argsh((x/a)) =ln((x/a)+(√(1+(x^2 /a^2 )))) ⇒ sh(4t) =(1/2){((x/a)+(√(1+(x^2 /a^2 ))))^4 −((x/a)+(√(1+(x^2 /a^2 ))))^(−4) } ⇒ I =((a^3 ∣a∣)/(8×32)){ ((x/a)+(√(1+(x^2 /a^2 ))))^4 −((x/a)+(√(1+(x^2 /a^2 ))))^(−4) } −((a^3 ∣a∣)/(32))ln((x/a)+(√(1+(x^2 /a^2 )))) +C .$
$l e t I = \int x^{2} \sqrt{a^{2} + x^{2}} d x c h a n g e m e n t x = a s h (t) g i v e$ $I = \int a^{2} {s h}^{2} (t) ∣ a ∣ c h (t) a c h (t) d t = a^{3} ∣ a ∣ \int {s h}^{2} (t) {c h}^{2} (t) d t$ $= a^{3} ∣ a ∣ \int {(\frac{c h (2 t) - 1}{2})}^{2} \times {(\frac{1 + c h (2 t)}{2})}^{2} d t$ $= \frac{a^{3} ∣ a ∣}{16} \int (c h (2 t) - 1) (c h (2 t) + 1) d t$ $= \frac{a^{3} ∣ a ∣}{16} \int ({c h}^{2} (2 t) - 1) d t = \frac{a^{3} ∣ a ∣}{16} \int (\frac{1 + c h (4 t)}{2} - 1) d t$ $= \frac{a^{3} ∣ a ∣}{32} \int (c h (4 t) - 1) d t$ $= \frac{a^{3} ∣ a ∣}{32 \times 4} s h (4 t) - \frac{a^{3} ∣ a ∣}{32} t + C$ $s h (4 t) = \frac{e^{4 t} - e^{- 4 t}}{2} a n d t = a r g s h (\frac{x}{a}) = l n (\frac{x}{a} + \sqrt{1 + \frac{x^{2}}{a^{2}}}) \Rightarrow$ $s h (4 t) = \frac{1}{2} {{(\frac{x}{a} + \sqrt{1 + \frac{x^{2}}{a^{2}}})}^{4} - {(\frac{x}{a} + \sqrt{1 + \frac{x^{2}}{a^{2}}})}^{- 4}} \Rightarrow$ $I = \frac{a^{3} ∣ a ∣}{8 \times 32} {{(\frac{x}{a} + \sqrt{1 + \frac{x^{2}}{a^{2}}})}^{4} - {(\frac{x}{a} + \sqrt{1 + \frac{x^{2}}{a^{2}}})}^{- 4}}$ $- \frac{a^{3} ∣ a ∣}{32} l n (\frac{x}{a} + \sqrt{1 + \frac{x^{2}}{a^{2}}}) + C .$
Commented by Power last updated on 20/Feb/20

$sir x = a tanu solution$
Commented by john santu last updated on 20/Feb/20

$\int x (x \sqrt{a^{2} + x^{2}}) d x = I$ $u = x \Rightarrow d u = d x$ $v = \int x \sqrt{a^{2} + x^{2}} d x = \frac{1}{2} \int \sqrt{a^{2} + x^{2}} d (a^{2} + x^{2})$ $= \frac{1}{3} {(a^{2} + x^{2})}^{\frac{3}{2}}$ $I = \frac{1}{3} x {(a^{2} + x^{2})}^{\frac{3}{2}} - \int {(a^{2} + x^{2})}^{\frac{3}{2}} d x$ $J = \int {(a^{2} + x^{2})}^{\frac{3}{2}} d x$ $l e t x = a \tan θ \Rightarrow d x = a \sec θ d θ$ $J = \int a^{2} \sec^{4} θ d θ$ $J = a^{2} \int (1 + \tan^{2} θ) d (\tan θ)$ $J = a^{2} (\tan θ + \frac{1}{3} \tan^{3} θ) + c$ $J = a^{2} (\frac{x}{a} + \frac{x^{3}}{3 a^{3}}) + c = a x + \frac{1}{3} \frac{x^{3}}{a} + c$ $∴ I - J = \frac{1}{3} \sqrt{{(a^{2} + x^{2})}^{3}} - a x - \frac{x^{3}}{3 a} + c$
Commented by Power last updated on 20/Feb/20

$thank you sir$
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