∫_0 ^( 1) t^2 (√(1+t^2 )) dt = ?


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Question Number 35379 by lilesh7 last updated on 18/May/18

$\int_{0}^{1} t^{2} \sqrt{1 + t^{2}} d t = ?$
Commented by prof Abdo imad last updated on 18/May/18
$let put I = ∫_0 ^1 t^2 (√(1+t^2 )) dt.changement t=shx give I = ∫_0 ^(ln(1+(√(2)))) sh^2 x chx.chxdx =∫_0 ^(ln(1+(√2))) (shx chx)^2 dx =(1/4) ∫_0 ^(ln(1+(√2))) sh^2 (2x)dx =(1/4) ∫_0 ^(ln(1+(√2))) ((ch(4x)−1)/2)dx =(1/8) ∫_0 ^(ln(1+(√2))) ch(4x)dx −(1/8)ln(1+(√2)) = (1/(32))[ sh(4x)]_0 ^(ln(1+(√2))) −(1/8)ln(1+(√2)) =(1/(32)) sh(4ln(1+(√(2)))) −(1/8)ln(1+(√2)) but we have shx =((e^x −e^(−x) )/2) ⇒sh(4ln(1+(√(2)))) =((e^(4ln(1+(√2))) − e^(−4ln(1+(√2))) )/2)=(((1+(√2))^4 −(1/((1+(√2))^4 )))/2)⇒ I = (1/(64)){ (1+(√2))^4 −(1+(√2))^(−4) } −(1/8)ln(1+(√2)) let remember that argsh(x)=ln(x+(√(x^2 +1))) argchx=ln(x +(√(x^2 −1)))$
$l e t p u t I = \int_{0}^{1} t^{2} \sqrt{1 + t^{2}} d t . c h a n g e m e n t t = s h x$ $g i v e I = \int_{0}^{l n (1 + \sqrt{2)}} {s h}^{2} x c h x . c h x d x$ $= \int_{0}^{l n (1 + \sqrt{2})} {(s h x c h x)}^{2} d x$ $= \frac{1}{4} \int_{0}^{l n (1 + \sqrt{2})} {s h}^{2} (2 x) d x = \frac{1}{4} \int_{0}^{l n (1 + \sqrt{2})} \frac{c h (4 x) - 1}{2} d x$ $= \frac{1}{8} \int_{0}^{l n (1 + \sqrt{2})} c h (4 x) d x - \frac{1}{8} l n (1 + \sqrt{2})$ $= \frac{1}{32} {[s h (4 x)]}_{0}^{l n (1 + \sqrt{2})} - \frac{1}{8} l n (1 + \sqrt{2})$ $= \frac{1}{32} s h (4 l n (1 + \sqrt{2)}) - \frac{1}{8} l n (1 + \sqrt{2}) b u t w e h a v e$ $s h x = \frac{e^{x} - e^{- x}}{2} \Rightarrow s h (4 l n (1 + \sqrt{2))}$ $= \frac{e^{4 l n (1 + \sqrt{2})} - e^{- 4 l n (1 + \sqrt{2})}}{2} = \frac{{(1 + \sqrt{2})}^{4} - \frac{1}{{(1 + \sqrt{2})}^{4}}}{2} \Rightarrow$ $I = \frac{1}{64} {{(1 + \sqrt{2})}^{4} - {(1 + \sqrt{2})}^{- 4}} - \frac{1}{8} l n (1 + \sqrt{2})$ $l e t r e m e m b e r t h a t$ $a r g s h (x) = l n (x + \sqrt{x^{2} + 1)}$ $a r g c h x = l n (x + \sqrt{x^{2} - 1})$
Commented by ajfour last updated on 18/May/18

$I = \frac{5 \sqrt{2}}{8} - \frac{\ln (1 + \sqrt{2})}{8} .$ $i s i t s a m e a s y o u r a n s w e r t o t h i s$ $q u e s t i o n, S i r ?$
Commented by abdo mathsup 649 cc last updated on 18/May/18

$p e r h a p s a f t e r d e v e l p p i n g t h e c a l c u l u s . . .$
	Answered by MJS last updated on 18/May/18

	$\int t^{2} \sqrt{t^{2} + 1} d t = \int (t^{2} + 1 - 1) \sqrt{t^{2} + 1} d t =$ $= \int ((t^{2} + 1) \sqrt{t^{2} + 1} - \sqrt{t^{2} + 1}) d t =$ $= \int {(t^{2} + 1)}^{\frac{3}{2}} d t - \int {(t^{2} + 1)}^{\frac{1}{2}} d t =$ $[t = \tan (u) \to d t = \sec^{2} (u) d u]$ $= \int \sec^{2} (u) {(\tan^{2} (u) + 1)}^{\frac{3}{2}} d u - \int \sec^{2} (u) {(\tan^{2} (u) + 1)}^{\frac{1}{2}} d u =$ $[\tan^{2} (u) + 1 = \sec^{2} (u)]$ $= \int \sec^{5} (u) d u - \int \sec^{3} (u) d u$ $now use the reduction formula$ $\int \sec^{n} (α) d α = \frac{\sec^{n - 2} (α) \tan (α)}{n - 1} + \frac{n - 2}{n - 1} \int \sec^{n - 2} (α) d α$
	Answered by sma3l2996 last updated on 18/May/18

	$I = \int_{0}^{1} t^{2} \sqrt{1 + t^{2}} d t = \int_{0}^{1} t \times t \sqrt{1 + t^{2}} d t$ $b y p a r t s$ $u = t \Rightarrow u^{'} = 1$ $v^{'} = t \sqrt{1 + t^{2}} \Rightarrow v = \frac{1}{3} (1 + t^{2}) \sqrt{1 + t^{2}}$ $S o I = \frac{1}{3} {[t (1 + t^{2}) \sqrt{1 + t^{2}}]}_{0}^{1} - \frac{1}{3} \int_{0}^{1} (1 + t^{2}) \sqrt{1 + t^{2}} d t$ $I = \frac{2 \sqrt{2}}{3} - \frac{1}{3} \int_{0}^{1} (\sqrt{1 + t^{2}} + t^{2} \sqrt{1 + t^{2}}) d t = \frac{2 \sqrt{2}}{3} - \frac{1}{3} \int_{0}^{1} \sqrt{1 + t^{2}} d t - \frac{1}{3} I$ $\frac{4}{3} I = \frac{2 \sqrt{2}}{3} - \frac{1}{3} \int_{0}^{1} \sqrt{1 + t^{2}} d t$ $l e t t = s i n h (u) \Rightarrow d t = c o s h (u) d u$ $\sqrt{1 + t^{2}} = \sqrt{1 + {s i n h}^{2} u} = c o s h (u)$ $S o \int_{0}^{1} \sqrt{1 + t^{2}} d t = \int_{0}^{{s i n h}^{- 1} (1)} {c o s h}^{2} (u) d u = \frac{1}{2} \int_{0}^{{s i n h}^{- 1} (1)} (1 + c o s h (2 u)) d u$ $= \frac{1}{2} {[u + \frac{1}{2} s i n h (2 u)]}_{0}^{{s i n h}^{- 1} (1)} = \frac{1}{2} ({s i n h}^{- 1} (1) + \sqrt{2})$ $= \frac{1}{2} (l n (1 + \sqrt{2}) + \sqrt{2})$ $S o I = \frac{\sqrt{2}}{2} - \frac{1}{4} (\frac{1}{2} l n (1 + \sqrt{2}) + \frac{\sqrt{2}}{2})$ $I = \frac{7 \sqrt{2}}{8} - \frac{1}{8} l n (1 + \sqrt{2})$
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