Missing some fundamental understanding- U-sub w/ radicals

[opus]

Homework Statement

The current section I'm working on has to do with arc length of a curve and surface area.
These all eventually end up with having to take the anti-derivative of a radical. At each instance, I get stuck by using u-substitution because when I take the derivative of $u$, my $du$ is not in the integrand and it's more than a scalar(it has operators or variables in it), so I can't pull them in front of the integral. Let me give an example, although I have the same problem for all of them.

Problem:
Find the length of the function of $x$ over the given interval. If you cannot evaluate the integral exactly, use technology to approximate it.
$y=\frac{1}{3}\left(x^2+2\right)^\frac{3}{2}$ from $x=0$ to $x=1$

Homework Equations

$Arc Length = \int_a^b \sqrt{1+\left[f'(x)\right]^2}dx$

The Attempt at a Solution

(i) $f'(x) = x\left(x^2+2\right)^{\frac{1}{2}}$
(ii) $\left[f'(x)\right]^2 = x^4+2x^2$
(iii) $Arc Length = \int_0^1 \sqrt{x^4+2x^2+1}dx$

This is where I always get stuck. No matter what I set $u$ equal to for u-substitution, $du$ is never left over in the integrand. It's always loaded with things that I can't pull out front of the integral. I tried using different forms of the expression under the radicand as well as not distributing $x^2\left(x^2+3\right)$ but to no avail.
I feel like there's some fundamental thing I am missing here because I run into the same problem every time.
Any ideas?

 

[pasmith]

Homework Statement

The current section I'm working on has to do with arc length of a curve and surface area.
These all eventually end up with having to take the anti-derivative of a radical. At each instance, I get stuck by using u-substitution because when I take the derivative of $u$, my $du$ is not in the integrand and it's more than a scalar(it has operators or variables in it), so I can't pull them in front of the integral. Let me give an example, although I have the same problem for all of them.

Problem:
Find the length of the function of $x$ over the given interval. If you cannot evaluate the integral exactly, use technology to approximate it.
$y=\frac{1}{3}\left(x^2+2\right)^\frac{3}{2}$ from $x=0$ to $x=1$

Homework Equations

$Arc Length = \int_a^b \sqrt{1+\left[f'(x)\right]^2}dx$

The Attempt at a Solution

(i) $f'(x) = x\left(x^2+2\right)^{\frac{1}{2}}$
(ii) $\left[f'(x)\right]^2 = x^4+2x^2$
(iii) $Arc Length = \int_0^1 \sqrt{x^4+2x^2+1}dx$

This is where I always get stuck. No matter what I set $u$ equal to for u-substitution, $du$ is never left over in the integrand. It's always loaded with things that I can't pull out front of the integral. I tried using different forms of the expression under the radicand as well as not distributing $x^2\left(x^2+3\right)$ but to no avail.
I feel like there's some fundamental thing I am missing here because I run into the same problem every time.
Any ideas?

The quantity under the square root is a perfect square: $x^4 + 2x^2 + 1 = (x^2 + 1)^2$.

In general, $$\int \sqrt{ax^2 + bx + c}\,dx = \int \sqrt{a\left(\left(x + \frac b{2a}\right)^2 - \frac{b^2 - 4ac}{4a^2}\right)}\,dx$$ calls for a substitution of the form $$x + \frac{b}{2a} = \left(\frac{|b^2 - 4ac|}{4a^2}\right)^{1/2} f(u)$$ where $f$ is a trigonometric or hyperbolic function depending on the signs of $a$ and $b^2 - 4ac$.

Square roots of higher order polynomials in general do not have antiderivatives in terms of elementary functions, so if you can't find a substitution which gives you something you can integrate then it may be that there isn't one.

 

[Ray Vickson]

Homework Statement

The current section I'm working on has to do with arc length of a curve and surface area.
These all eventually end up with having to take the anti-derivative of a radical. At each instance, I get stuck by using u-substitution because when I take the derivative of $u$, my $du$ is not in the integrand and it's more than a scalar(it has operators or variables in it), so I can't pull them in front of the integral. Let me give an example, although I have the same problem for all of them.

Problem:
Find the length of the function of $x$ over the given interval. If you cannot evaluate the integral exactly, use technology to approximate it.
$y=\frac{1}{3}\left(x^2+2\right)^\frac{3}{2}$ from $x=0$ to $x=1$

Homework Equations

$Arc Length = \int_a^b \sqrt{1+\left[f'(x)\right]^2}dx$

The Attempt at a Solution

(i) $f'(x) = x\left(x^2+2\right)^{\frac{1}{2}}$
(ii) $\left[f'(x)\right]^2 = x^4+2x^2$
(iii) $Arc Length = \int_0^1 \sqrt{x^4+2x^2+1}dx$

This is where I always get stuck. No matter what I set $u$ equal to for u-substitution, $du$ is never left over in the integrand. It's always loaded with things that I can't pull out front of the integral. I tried using different forms of the expression under the radicand as well as not distributing $x^2\left(x^2+3\right)$ but to no avail.
I feel like there's some fundamental thing I am missing here because I run into the same problem every time.
Any ideas?

As pasmith shows, you are lucky in this case. However, in general, integrands of the form $\sqrt{x^3 + a x^2+bx +c}, \sqrt{x^4 + a x^3 + b x^2 + cx + d}$ or $\sqrt{(x^2+ax+b)(x^2+cx+d)}$ are provably non-elementary, but need "elliptic functions" when integrating. By "provably" I mean that it has been shown (as a rigorous theorem with an iron-clad proof) that it is impossible to write a formula for such integrals in terms of a finite number of elementary functions (powers, roots, trig or inverse trig, exponential, logarithmic, etc.). Even if you allow yourself a billion pages to write out a formula, that would not be enough: you would need infinitely many pages!

 

[opus]

The quantity under the square root is a perfect square: $x^4 + 2x^2 + 1 = (x^2 + 1)^2$.

In general, $$\int \sqrt{ax^2 + bx + c}\,dx = \int \sqrt{a\left(\left(x + \frac b{2a}\right)^2 - \frac{b^2 - 4ac}{4a^2}\right)}\,dx$$ calls for a substitution of the form $$x + \frac{b}{2a} = \left(\frac{|b^2 - 4ac|}{4a^2}\right)^{1/2} f(u)$$ where $f$ is a trigonometric or hyperbolic function depending on the signs of $a$ and $b^2 - 4ac$.

Square roots of higher order polynomials in general do not have antiderivatives in terms of elementary functions, so if you can't find a substitution which gives you something you can integrate then it may be that there isn't one.

As pasmith shows, you are lucky in this case. However, in general, integrands of the form $\sqrt{x^3 + a x^2+bx +c}, \sqrt{x^4 + a x^3 + b x^2 + cx + d}$ or $\sqrt{(x^2+ax+b)(x^2+cx+d)}$ are provably non-elementary, but need "elliptic functions" when integrating. By "provably" I mean that it has been shown (as a rigorous theorem with an iron-clad proof) that it is impossible to write a formula for such integrals in terms of a finite number of elementary functions (powers, roots, trig or inverse trig, exponential, logarithmic, etc.). Even if you allow yourself a billion pages to write out a formula, that would not be enough: you would need infinitely many pages!

Oh wow that's pretty intense. There seems to be way more intricacies with anti-derivatives than I had originally thought.
And it looks like I need to work on my factoring skills a bit more for these.
So in terms of writing the square root to the power of 1/2, and using the reverse power rule, does this not work on expressions? For example, reversing the chain rule by setting the polynomial to $u$, then using reverse power rule?

 

[pasmith]

Oh wow that's pretty intense. There seems to be way more intricacies with anti-derivatives than I had originally thought.
And it looks like I need to work on my factoring skills a bit more for these.
So in terms of writing the square root to the power of 1/2, and using the reverse power rule, does this not work on expressions? For example, reversing the chain rule by setting the polynomial to $u$, then using reverse power rule?

Setting $u = f(x)$ in $\int \sqrt{f(x)}\,dx$ yields $$\int \frac{\sqrt{u} }{f'(f^{-1}(u))}\,du$$ which may or may not be easier to integrate than the original. Had you started with $\int \sqrt{f(x)} f'(x)\,dx$ then setting $u = f(x)$ would yield $\int \sqrt{u}\,du = \frac{2}{3}u^{3/2} + C$.

 

[Mark44]

There seems to be way more intricacies with anti-derivatives than I had originally thought.

Finding an antiderivative is waaaaay more complicated than finding a derivative. For differentiation, we have a basic definition (limit of difference quotient) and a set of rules that stem from the definition. For integration, it's a little like Jeopardy -- you have the answer (an integrand function) and need to come up with the question (what other function can be differentiated to yield the integrand).

So in terms of writing the square root to the power of 1/2, and using the reverse power rule, does this not work on expressions? For example, reversing the chain rule by setting the polynomial to u, then using reverse power rule?

Ordinary substitution ("u" substitution) is the reverse of the chain rule. I don't know if you have been exposed to integration by parts yet. This integration technique is the reverse of the product rule in differentiation.

 

[opus]

Finding an antiderivative is waaaaay more complicated than finding a derivative. For differentiation, we have a basic definition (limit of difference quotient) and a set of rules that stem from the definition. For integration, it's a little like Jeopardy -- you have the answer (an integrand function) and need to come up with the question (what other function can be differentiated to yield the integrand).

Ordinary substitution ("u" substitution) is the reverse of the chain rule. I don't know if you have been exposed to integration by parts yet. This integration technique is the reverse of the product rule in differentiation.

I'm beginning to notice this. Some integrands that look extremely simple have had me sweating bullets for some time until I just used a computer to see the anti-derivative which turned out to be a huge strung out complicated thing. I've been starting to memorize some useful identities to see if I can just rewrite things into different terms, but when radicals are involved, it usually doesn't seem to matter how I write them. One problem told us to evaluate the integral with a computer but I wanted to try it on my own and and couldn't get it so I went to a tutor to see how to do it and he took up a page to find the anti-derivative of what seemed like a harmless one and he told me about integration by parts.
I think this is something I will learn in Calculus II next semester. We are just finishing up our last topic and it has to do with intro to differential equations and initial value problems.

 

[fresh_42]

Finding an antiderivative is waaaaay more complicated than finding a derivative.

My mathematics teacher used to say: Everybody can differentiate, but only artists can integrate, too.

 

[opus]

My mathematics teacher used to say: Everybody can differentiate, but only artists can integrate, too.

I like that!