# Complicated Integral Using the Substitution Method

• Maburo
In summary, it seems that the integral cannot be evaluated using a simple u substitution. Further simplification produces an equation in terms of cosine and tangent functions.f

## Homework Statement

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

## Homework Equations

If $$f(x)=g(u(x))u'(x)$$
and $$\int g(x)dx = G(x) +C$$
then $$\int f(x)dx = G(u(x))+C$$

## The Attempt at a Solution

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

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!

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.

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.

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.

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 $$\int\frac{dx}{\sqrt{1-\sin^4(x)}} = \int\frac{dx}{\cos^2(x)\sqrt{1+2\tan^2(x)}}$$
An obvious substitution for this would then be $$u=tan(x) \implies \int\frac{dx}{\cos^2(x)\sqrt{1+2\tan^2(x)}} = \int\frac{du}{\sqrt{1+2u^2}}$$
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.

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.
$$\int\frac{dx}{\sqrt{1-\sin^4(x)}} = \frac{\sqrt{2}}{4}\ln\Biggr\vert 4\tan^2(x)+2\tan(x)\sqrt{2+4\tan^2(x)}+1 \Biggr\vert + C$$
I was pretty careful during the calculation, so I hope I have not made an error. Thanks again for the help!

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.
$$\int\frac{dx}{\sqrt{1-\sin^4(x)}} = \frac{\sqrt{2}}{4}\ln\Biggr\vert 4\tan^2(x)+2\tan(x)\sqrt{2+4\tan^2(x)}+1 \Biggr\vert + C$$
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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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.