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Complicated Integral Using the Substitution Method

  1. Jan 19, 2015 #1
    1. The problem statement, all variables and given/known data
    Evaluate the following integral using a change of variables:
    [tex]\int\frac{dx}{\sqrt{1-\sin^4{x}}}[/tex]

    2. Relevant equations
    If [tex]f(x)=g(u(x))u'(x)[/tex]
    and [tex]\int g(x)dx = G(x) +C [/tex]
    then [tex]\int f(x)dx = G(u(x))+C [/tex]

    3. The attempt at a solution
    It seems helpful to first simplify a little to obtain [tex]\int\frac{dx}{\sqrt{1-\sin^4(x)}} = \int\frac{dx}{cos(x)\sqrt{1+\sin^2(x)}}[/tex]
    From this, further simplification produces, [tex]\sqrt{\frac{2}{3}}\int\frac{dx}{cos(x)\sqrt{1-\frac{1}{3}\cos(2x)}}[/tex] from which I cannot determine a useful change of variable.
    On another attempt, using some substitutions (leaving my work out), I obtained [tex]\frac{1}{2}\int\frac{du}{(2-u)\sqrt{u-1}\sqrt{u}}[/tex]

    Hopefully I have not made any errors in my calculations. I cannot find a useful substitution from any of these steps. Is there any trick or further simplification that can be made in order to make this easier to evaluate? Thanks for the help!
     
  2. jcsd
  3. Jan 19, 2015 #2

    LCKurtz

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    I don't think this is an elementary integral that can be done with a simple u substitution. Maple gives a rather complicated expression in terms of Elliptic functions.
     
  4. Jan 19, 2015 #3

    Dick

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    Maxima gives a reasonably simple answer in terms of elementary functions. Looking at the answer I can see how the substitution works but I don't see any direct way to guess it.
     
  5. Jan 19, 2015 #4

    RUber

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    What if you were to let u = sin^2 x, then du = 2 cos x sin x , giving you
    ## \int \frac{ du }{2\sqrt{ (1- u) ( u )( 1-u^2)}}=\int \frac{ du }{2\sqrt{ u^4-u^3-u^2+u }}##
    This is a polynomial to the (-1/2) power, and can be evaluated by partial fractions.
    Again, I second what LCKurtz said above, this does not give a pretty answer.
     
  6. Jan 19, 2015 #5

    Dick

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    Here's a hint to derive a pretty answer. Take a factor out of the square root so you get a cos^2(x) in the denominator. Why? If you rearrange what's left under the square root, you'll see.
     
    Last edited: Jan 19, 2015
  7. Jan 19, 2015 #6
    If I am understanding correctly, you mean to write [tex]\int\frac{dx}{\sqrt{1-\sin^4(x)}} = \int\frac{dx}{\cos^2(x)\sqrt{1+2\tan^2(x)}}[/tex]
    An obvious substitution for this would then be [tex]u=tan(x) \implies \int\frac{dx}{\cos^2(x)\sqrt{1+2\tan^2(x)}} = \int\frac{du}{\sqrt{1+2u^2}}[/tex]
    This is much easier to work with. I think I can find a solution to this now. Thanks a lot for the hint, Dick! It really helped. A pretty answer is near, indeed.
     
  8. Jan 19, 2015 #7
    After a few more substitutions, a few algebra tricks, and a lot of simplification, I think I have gotten the answer. It required only change of variables.
    [tex]\int\frac{dx}{\sqrt{1-\sin^4(x)}} = \frac{\sqrt{2}}{4}\ln\Biggr\vert 4\tan^2(x)+2\tan(x)\sqrt{2+4\tan^2(x)}+1 \Biggr\vert + C [/tex]
    I was pretty careful during the calculation, so I hope I have not made an error. Thanks again for the help!
     
  9. Jan 19, 2015 #8

    Dick

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    Probably right, I can't check it right now. It comes out even nicer using the arcsinh function if you know inverse hyperbolics. (arcsinh(x))'=1/sqrt(1+x^2).

    EDIT: Trying to check it... I don't think that looks right.
     
    Last edited: Jan 20, 2015
  10. Jan 20, 2015 #9

    LCKurtz

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    Maple users may find it interesting that if you give it the problem in this form:
    $$\int\frac{dx}{\cos^2(x)\sqrt{1+2\tan^2(x)}}$$it cranks out an indecipherable answer including elliptic functions whereas if you make the ##u = tan(x)## substitution and give it this form$$
    \int\frac{du}{\sqrt{1+2u^2}}$$it cranks out the nice arcsinh form. I guess that shows that Dick is smarter than Maple.:oldsmile:
     
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