calculate ∫_1 ^(√2) (x^3 +1)(√(x^2 −1))dx


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Question Number 54919 by maxmathsup by imad last updated on 14/Feb/19

$c a l c u l a t e \int_{1}^{\sqrt{2}} (x^{3} + 1) \sqrt{x^{2} - 1} d x$
Commented by Abdo msup. last updated on 15/Feb/19
$let I =∫_1 ^(√2) (x^3 +1)(√(x^2 −1)) dx changement x=cht gove I =∫_(argch(1)) ^(argch((√2))) (ch^3 t+1)sh^2 (t) dt =∫_0 ^(ln(1+(√2))) ch^3 t sh^2 t dt +∫_0 ^(ln(1+(√2))) sh^2 t dt ∫_0 ^(ln(1+(√2))) ((ch(2t)−1)/2)dt =(1/4)[sh(2t)]_0 ^(ln(1+(√2))) −((ln(1+(√2)))/2) =(1/8)[ e^(2t) −e^(−2t) ]_0 ^(ln(1+(√2))) −((ln(1+(√2)))/2) =(1/8){ (1+(√2))^2 −(1/((1+(√2))^2 ))}−((ln(1+(√2)))/2) also we have ch^3 tsh^2 t =(chtsht)^2 cht =(1/4)sh^2 (2t)ch(t) =(1/4)(((ch(4t)−1)/2))ch(t) =(1/8)(((e^(4t) +e^(−4t) )/2) −1)(((e^t +e^(−t) )/2)) =(1/(32))(e^(4t) +e^(−4t) −2)(e^t +e^(−t) ) =(1/(32))(e^(5t ) +e^(3t ) +e^(−3t) +e^(−5t) −2e^t −2e^(−t) ) ⇒ ∫_0 ^(ln(1+(√2))) ch^3 t sh^2 t dt =(1/(32)) ∫_0 ^(ln(1+(√2))) ( e^(5t) +e^(−5t) +e^(3t) +e^(−3t) −2e^t −2e^(−t) )dt =(1/(32))[(1/5)e^(5t) −(1/5)e^(−5t) +(1/3)e^(3t) −(1/3)e^(−3t) −2e^t +2e^(−t) ]_0 ^(ln(1+(√2))) =(1/(32)){ (1/5)( (1+(√2))^2 −(1/((1+(√2))^2 )))+(1/3)((1+(√2))^3 −(1/((1+(√2))^3 ))) −2( 1+(√2)−(1/(1+(√2))))} so the value of I is determined$
$l e t I = \int_{1}^{\sqrt{2}} (x^{3} + 1) \sqrt{x^{2} - 1} d x c h a n g e m e n t x = c h t g o v e$ $I = \int_{a r g c h (1)}^{a r g c h (\sqrt{2})} ({c h}^{3} t + 1) {s h}^{2} (t) d t$ $= \int_{0}^{l n (1 + \sqrt{2})} {c h}^{3} t {s h}^{2} t d t + \int_{0}^{l n (1 + \sqrt{2})} {s h}^{2} t d t$ $\int_{0}^{l n (1 + \sqrt{2})} \frac{c h (2 t) - 1}{2} d t = \frac{1}{4} {[s h (2 t)]}_{0}^{l n (1 + \sqrt{2})}$ $- \frac{l n (1 + \sqrt{2})}{2} = \frac{1}{8} {[e^{2 t} - e^{- 2 t}]}_{0}^{l n (1 + \sqrt{2})} - \frac{l n (1 + \sqrt{2})}{2}$ $= \frac{1}{8} {{(1 + \sqrt{2})}^{2} - \frac{1}{{(1 + \sqrt{2})}^{2}}} - \frac{l n (1 + \sqrt{2})}{2}$ $a l s o w e h a v e {c h}^{3} {t s h}^{2} t = {(c h t s h t)}^{2} c h t$ $= \frac{1}{4} {s h}^{2} (2 t) c h (t) = \frac{1}{4} (\frac{c h (4 t) - 1}{2}) c h (t)$ $= \frac{1}{8} (\frac{e^{4 t} + e^{- 4 t}}{2} - 1) (\frac{e^{t} + e^{- t}}{2})$ $= \frac{1}{32} (e^{4 t} + e^{- 4 t} - 2) (e^{t} + e^{- t})$ $= \frac{1}{32} (e^{5 t} + e^{3 t} + e^{- 3 t} + e^{- 5 t} - 2 e^{t} - 2 e^{- t}) \Rightarrow$ $\int_{0}^{l n (1 + \sqrt{2})} {c h}^{3} t {s h}^{2} t d t$ $= \frac{1}{32} \int_{0}^{l n (1 + \sqrt{2})} (e^{5 t} + e^{- 5 t} + e^{3 t} + e^{- 3 t} - 2 e^{t} - 2 e^{- t}) d t$ $= \frac{1}{32} {[\frac{1}{5} e^{5 t} - \frac{1}{5} e^{- 5 t} + \frac{1}{3} e^{3 t} - \frac{1}{3} e^{- 3 t} - 2 e^{t} + 2 e^{- t}]}_{0}^{l n (1 + \sqrt{2})}$ $= \frac{1}{32} {\frac{1}{5} ({(1 + \sqrt{2})}^{2} - \frac{1}{{(1 + \sqrt{2})}^{2}}) + \frac{1}{3} ({(1 + \sqrt{2})}^{3} - \frac{1}{{(1 + \sqrt{2})}^{3}})$ $- 2 (1 + \sqrt{2} - \frac{1}{1 + \sqrt{2}})} s o t h e v a l u e o f I i s d e t e r m i n e d$
	Answered by tanmay.chaudhury50@gmail.com last updated on 14/Feb/19

	$\int x \times x^{2} \sqrt{x^{2} - 1} d x + \int \sqrt{x^{2} - 1} d x$ $t^{2} = x^{2} - 1 \to t d t = x d x$ $\int (t^{2} + 1) t \times t d t + \int \sqrt{x^{2} - 1} d x$ $\int t^{4} + t^{2} d t + \int \sqrt{x^{2} - 1} d x$ $\frac{t^{5}}{5} + \frac{t^{3}}{3} + \frac{x}{2} \sqrt{x^{2} - 1} - \frac{1}{2} l n (x + \sqrt{x^{2} - 1}) + c$ $\frac{{(x^{2} - 1)}^{\frac{5}{2}}}{5} + \frac{{(x^{2} - 1)}^{\frac{3}{2}}}{3} + \frac{x}{2} \sqrt{x^{2} - 1} - \frac{1}{2} l n (x + \sqrt{x^{2} - 1}) + c$ $s o a n s w e r i s$ $∣ \frac{{(x^{2} - 1)}^{\frac{5}{2}}}{5} + \frac{{(x^{2} - 1)}^{\frac{3}{2}}}{3} + \frac{x}{2} \sqrt{x^{2} - 1} - \frac{1}{2} l n (x + \sqrt{x^{2} - 1}) ∣_{1}^{\sqrt{2}}$ $[\frac{1}{5} + \frac{1}{3} + \frac{\sqrt{2}}{2} \sqrt{2 - 1} - \frac{1}{2} l n (\sqrt{2} + 1)]$ $\frac{8}{15} + \frac{\sqrt{2}}{2} - \frac{1}{2} l n (\sqrt{2} + 1)$
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