Complicated Integral Using the Substitution Method

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Homework Statement


Evaluate the following integral using a change of variables:
[tex]\int\frac{dx}{\sqrt{1-\sin^4{x}}}[/tex]

Homework 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]

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!
 

Answers and Replies

  • #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.
 
  • #3
Dick
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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.
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.
 
  • #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.
 
  • #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.
 
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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.
 
  • #7
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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!
 
  • #8
Dick
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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!
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.
 
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  • #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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