∫_(−π/2) ^(π/2) sin {log (x+(√(x^2 +1)) )} dx =


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Question Number 34446 by dilanho last updated on 06/May/18
$∫_(−π/2) ^(π/2) sin {log (x+(√(x^2 +1)) )} dx =$
$\underset{- π / 2}{\int^{π / 2}} \sin {\log (x + \sqrt{x^{2} + 1})} d x =$
Commented by math khazana by abdo last updated on 06/May/18
$let put I = ∫_(−(π/2)) ^(π/2) sin{ln(x +(√(x^2 +1)) )}dx I = ∫_(−(π/2)) ^(π/2) sin(argshx)dx let put argshx=t ⇔ x=sh(t) ⇒ I = ∫_(−argsh((π/2))) ^(argsh((π/2))) sint cht dt I =0 because the function t →sint cht is odd.$
$l e t p u t I = \int_{- \frac{π}{2}}^{\frac{π}{2}} s i n {l n (x + \sqrt{x^{2} + 1})} d x$ $I = \int_{- \frac{π}{2}}^{\frac{π}{2}} s i n (a r g s h x) d x l e t p u t a r g s h x = t \Leftrightarrow$ $x = s h (t) \Rightarrow I = \int_{- a r g s h (\frac{π}{2})}^{a r g s h (\frac{π}{2})} s i n t c h t d t$ $I = 0 b e c a u s e t h e f u n c t i o n t \to s i n t c h t i s o d d .$
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