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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 =
$$\:\underset{−\pi/\mathrm{2}} {\overset{\pi/\mathrm{2}} {\int}}\:\mathrm{sin}\:\left\{\mathrm{log}\:\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:\right)\right\}\:{dx}\:= \\ $$
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.
$${let}\:{put}\:{I}\:=\:\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:{sin}\left\{{ln}\left({x}\:+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:\right)\right\}{dx} \\ $$$${I}\:\:=\:\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:{sin}\left({argshx}\right){dx}\:\:{let}\:{put}\:{argshx}={t}\:\Leftrightarrow \\ $$$${x}={sh}\left({t}\right)\:\Rightarrow\:{I}\:\:=\:\int_{−{argsh}\left(\frac{\pi}{\mathrm{2}}\right)} ^{{argsh}\left(\frac{\pi}{\mathrm{2}}\right)} \:{sint}\:\:{cht}\:{dt} \\ $$$${I}\:\:=\mathrm{0}\:{because}\:{the}\:{function}\:{t}\:\rightarrow{sint}\:{cht}\:{is}\:{odd}. \\ $$

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