Question

\[\int\limits_0^{1/\sqrt{3}} \sqrt{x+ \sqrt{x^2+1}}dx\]

Answer 1

i would start with a traditional approach,
x = tan y

anyone with smarter solution?

Answer 2

i think
\(u = \sqrt{x+ \sqrt{x^2+1}}\)
will work better! :)

Answer 3

could you give further hints into the problem? we only have three minutes to solve these.

Answer 4

http://math.mit.edu/~sswatson/pdfs/qualifying_round_2014.pdf

Answer 5

Let \(x=\sinh y\), the integral becomes
\[ \int\limits_{..}^{..}\sqrt{\sinh y+ \cosh y}~ \cosh y\, dy = \int\limits_{..}^{..}\sqrt{e^y}~ \cosh y\, dy \]

Answer 6

replace \(\cosh y\) by \(\dfrac{e^y+e^{-y}}{2}\) and rest is trivial

Answer 7

please try this substitution:
\[\large \sqrt {{x^2} + 1} = x + t\]
where \(t\) is the new variable

Answer 8

is that euler's substitution or something like that?

Answer 9

ganesh, excellent thought!
would substituting the limits be a challenge in your approach?!

Answer 10

\[x=0\implies y=\arcsin h (0) = 0\]
\[x=1/\sqrt{3}\implies y=\arcsin h (1/\sqrt{3}) = \ln(1/\sqrt{3} + \sqrt{1/3+1}) = \ln(\sqrt{3})\]

Answer 11

\(x=\sinh y =\dfrac{e^y-e^{-y}}{2}\)

\(e^{2y}-2xe^y-1=0 \)

\(y = \arcsin h(x)= \ln(x+\sqrt{x^2+1})\)

Answer 12

Great work!

Answer 13

Ofcourse this hyperbolic substitution saves time only if you're familiar with that identity

Answer 14

http://sma.epfl.ch/cours/csma/Analysis1/Images/analysis1-2014-15876x.png

Answer 15

if we know
\(\sinh y = \ln (x+\sqrt{x^2+1 })\)

then we could easily think of modifying our original integral function to
\(\large \sqrt{e^{\arcsin h \: x}} \)

in that case, the substitution y = arcsinh x
or x = sinh y would become very obvious...