The Tricky Antiderivative: Solving the Integral of a Radical Function

[pierce15]

1. The problem, the whole problem, and nothing but the problem

$$ \int \sqrt{\frac{1+x^2}{1-x^2}} dx $$

Homework Equations

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The Attempt at a Solution

I tried a partial fraction decomp, but this obviously didn't get me very far because this isn't a rational function. Is there some kind of ingenious substitution applicable here?

 

[Dick]

I don't see any way to get an antiderivative. And Wolfram Alpha says it's an elliptic integral. I don't disagree. If it's a definite integral there might be a trick. Is that the whole problem?

 

[pierce15]

I don't see any way to get an antiderivative. And Wolfram Alpha says it's an elliptic integral. I don't disagree. If it's a definite integral there might be a trick. Is that the whole problem?

... I made it up

I've changed the problem to the following:

$$ \int \sqrt{\frac{1+x}{1-x}} dx $$

According to wolfram, this does have a (nasty) elementary antiderivative. Any thoughts?

 

[Dick]

... I made it up

You are making up HARD PROBLEMS.

 

[pierce15]

Sorry for adding new detail to a previous post by the way, I just remembered that that's a big no-no

 

[iRaid]

Making your own integrals isn't a good idea because you tend not to know if you can solve them. For example: $\int e^{x^{2}} dx$

 

[runningninja]

The new integral you made up is solvable however. Do a u substitution.

 

[Curious3141]

... I made it up

I've changed the problem to the following:

$$ \int \sqrt{\frac{1+x}{1-x}} dx $$

According to wolfram, this does have a (nasty) elementary antiderivative. Any thoughts?

Substitute something that would make that square root on the denominator much easier to deal with (or eliminate).

 

[pierce15]

How about the substitution $x = -2u - u^2$, $dx = -2 - 2u$

$$ \int \sqrt{ \frac{1 - 2u-u^2}{1+2u + u^2} } (-2-2u) \, du $$

$$ -2 \int \frac{ \sqrt{1-2u-u^2} }{1+u}(1+u) \, du $$

$$ -2 \int \sqrt{1-2u-u^2} \, du$$

Alright, that looks fairly simple, but I'm blanking as to how I should deal with it.

 

[Dick]

I'd say complete the square and then use a trig substitution.

 

[Curious3141]

How about the substitution $x = -2u - u^2$, $dx = -2 - 2u$

$$ \int \sqrt{ \frac{1 - 2u-u^2}{1+2u + u^2} } (-2-2u) \, du $$

$$ -2 \int \frac{ \sqrt{1-2u-u^2} }{1+u}(1+u) \, du $$

$$ -2 \int \sqrt{1-2u-u^2} \, du$$

Alright, that looks fairly simple, but I'm blanking as to how I should deal with it.

Why not $x = 1 - u^2$?

 

[pierce15]

Alright, I completed the square, changing $1-2u-u^2$ to $2-(1+u)^2$.

$$ -2\int \sqrt{2-(1+u)^2} \, du $$

$$1+u = \sqrt{2} \sin\theta, \quad du = \sqrt{2} \cos \theta \, d\theta $$

$$ -2\sqrt{2} \int \cos \theta \sqrt{2-2 \sin^2\theta} \, d\theta $$

$$ -4 \int \cos \theta \sqrt{1-\sin^2\theta} \, d\theta $$

$$ v = \sin \theta, \quad dv = \cos \theta \, d\theta$$

$$ -4 \int 1 - v^2 \, dv $$

$$ -4( v - v^3/3) $$

$$ 4(v^3/3 - v) $$

Before I go back and perform the various substitutions, is that correct?

 

[pierce15]

Why not $x = 1 - u^2$?

That would have truly been a wise choice

 

[runningninja]

Alright, I completed the square, changing $1-2u-u^2$ to $2-(1+u)^2$.

$$ -2\int \sqrt{2-(1+u)^2} \, du $$

$$1+u = \sqrt{2} \sin\theta, \quad du = \sqrt{2} \cos \theta \, d\theta $$

$$ -2\sqrt{2} \int \cos \theta \sqrt{2-2 \sin^2\theta} \, d\theta $$

$$ -4 \int \cos \theta \sqrt{1-\sin^2\theta} \, d\theta $$

$$ v = \sin \theta, \quad dv = \cos \theta \, d\theta$$

$$ -4 \int 1 - v^2 \, dv $$

$$ -4( v - v^3/3) $$

$$ 4(v^3/3 - v) $$

Before I go back and perform the various substitutions, is that correct?

Isn't 1 - sin²θ just cos²θ?

 

[pierce15]

Isn't 1 - sin²θ just cos²θ?

-.- yes

$$ -4 \int \cos \theta \sqrt{1-\sin^2\theta} \, d\theta $$
$$ -4 \int \cos^2 \theta \, d\theta $$
$$ -4 \int \frac{1+\cos(2\theta)}{2} d\theta $$
$$ -2\Big(\theta + \int \cos(2\theta) d\theta \Big)$$
$$ -2(\theta + \frac{ \sin(2\theta)}{2}) $$
$$ - \sin(2\theta) -2\theta $$