Integral of arcsin (x^2)

  • Thread starter Kiwiro0ls
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  • #1
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Just got out of a Calculus 2/ Lineal Algebra exam and im still wondering about the first question:
∫arcsin(2×2)dx

I used integration by parts letting u=arcsin(2×2) and so du=dx/sqrt[1-4x4], v=x

=xarcsin(2×2) - ∫[x/sqrt[1-4x4]dx

this second integral is what I can´t solve.

I tried doing trigonometric substitution where sin∅=2x2, and then later used a trig identity to integrate... but I really think I messed up somewhere.

If anyone out there has any clue, its much appreciated! :)
 
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  • #2
If the second integral WERE ∫[x/sqrt[1-4x^4]dx, you could just substitute w=x^2 which would then give you arcsin again.
However, du=dx 4x/sqrt[1-4x4] (chain rule) so that the integral you actually have to solve becomes ∫[4x^2/sqrt[1-4x^4]dx which, according to wolfram alpha has a complicated solution involving elliptic integrals.
Maybe you are missing a factor in the question?
 
  • #3
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No im very very very sure this is the question,
theres some talk amongst the class that it was typed wrong or something, apparently no one got those 15pts...
anyway, im still out of ideas and trying at it!
 
  • #4
SammyS
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Just got out of a Calculus 2/ Lineal Algebra exam and im still wondering about the first question:
∫arcsin(2×2)dx

I used integration by parts letting u=arcsin(2×2) and so du=dx/sqrt[1-4x4], v=x

=xarcsin(2×2) - ∫[x/sqrt[1-4x4]dx

this second integral is what I can´t solve.

I tried doing trigonometric substitution where sin∅=2x2, and then later used a trig identity to integrate... but I really think I messed up somewhere.

If anyone out there has any clue, its much appreciated! :)
Are you sure it wasn't something like [itex]\displaystyle \int x\ \arcsin(2x^2)\,dx\ ?[/itex]
 
  • #5
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Nope D: im 100% sure it was ∫arcsin (2x^2)dx
 
  • #6
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Nope D: im 100% sure it was ∫arcsin (2x^2)dx

Err I'm surprised a question like that would show up, but you can use the fact that :

[itex]\int_{}^{}arcsin(2x^2)dx = \int_{}^{}\frac{\pi}{2} - arccosxdx = \frac{\pi}{2}x - xarccosx + \sqrt{1-x^2} + c[/itex]
 
  • #7
LCKurtz
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Err I'm surprised a question like that would show up, but you can use the fact that :

[itex]\int_{}^{}arcsin(2x^2)dx = \int_{}^{}\frac{\pi}{2} - arccosxdx = \frac{\pi}{2}x - xarccosx + \sqrt{1-x^2} + c[/itex]

Wouldn't it be ##\int \arcsin(2x^2)\, dx = \int\frac \pi 2 -\arccos(2x^2)\, dx##?
 
  • #8
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So
=xarcsin(2x2)-4∫[x2dx/sqrt(1-4x4)]
and thats what im trying to solve
 
  • #9
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Looks like a valid problem for trigonometric substitution, particularly the 1/sqrt(...).
 
  • #10
LCKurtz
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Beats me why this thread is still alive. As has already been pointed out, the solution involves elliptic functions and no "standard" calculus technique is going to work it. It is pretty obvious that if that was intended, the problem has a typo.
 
  • #11
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oh well :´(
 

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