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I am attempting to learn about the substitution rule and I can't make sense of what Stewart is trying to say. "To find this integral we use the problem solving strategy of introducing something extra. Here the something extra is a new variable, we change from the variable x to a new variable u. Suppose that we let u be the quantity under the root sing 1, $u=1+x^2$ Then the differential of u is du=2xdx. Notice that if the dx in the notation for an integral were to be interpreted as a differnetial then the differential 2xdx would occur in 1 and so formally without justifying our calculation we could write" The rest doesn't really matter, I just don't understand what is going on at all.

The differential of u is $.5(1+x^2)(2x)$ not what he has.

1: $\int 2x \sqrt{1+x^2}$

Arturo Magidin
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  • How do you get that the differential of $u$ is $.5(1+x^2)2x$? If $u=1+x^2$, then $du = d(1+x^2) = d(1)+d(x^2) = 0 + 2x,dx$, exactly what Stewart says. – Arturo Magidin Nov 08 '11 at 17:50
  • What does du=2xdx mean? The differnetial of du with respect to u is equal to 2x differential of dx with respect to x? –  Nov 08 '11 at 17:58
  • It means that if $u=1+x^2$, then "a small change in $u$" (the differential) will necessarily be the same size as a "small change in $x$" (the differential in $x$) times twice the value of $x$ you are in. But I repeat my question: why is it that you think that "the differential of $u$ is $0.5(1+x^2)(2x)$." In your mind, what does that mean and why is that what the differential of $u$ "is"? – Arturo Magidin Nov 08 '11 at 18:06
  • I was reading it too quickly and took the differential of the square root of u not u. I really do not know what you are talking about small changes, is this something to do with reimann sums? –  Nov 08 '11 at 18:20
  • Do you remember what a differential is? If you don't know what a differential is, then of course you can't make sense of any explanation for why they are what they are. I don't know if this will help. – Arturo Magidin Nov 08 '11 at 18:27
  • Off the top of my head a differential is the small difference in a function. –  Nov 08 '11 at 18:57
  • I have read that already and I can make sense of it a little bit but I always forget. –  Nov 08 '11 at 19:16
  • @Jordan Might I suggest you take a look at The Calculus Lifesaver videos - http://press.princeton.edu/video/banner/ - specifically video 12 addresses u-substitution. – mboratko Nov 09 '11 at 18:37

2 Answers2

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The $u$ he is referring to is $1+x^2$, not $\sqrt{1+x^2}$.

I prefer the following explanation for ``$u$-substitution'' (which avoids any $u$'s at all):

The basic method for finding an antiderivative is to recall a corresponding derivative rule: if you know a basic derivative rule, then you know a basic antiderivative rule.

Here, we regard a function arising from an application of the chain rule as a ``basic derivative''. Given one of these, it is easy to find its antiderivative so long as you recognize that it is a derivative resulting from using the chain rule.

What expressions result from applying the chain rule? Well, they have the form of a product of a composition of functions with the derivative of the inner function of the composition.

For example, for the integral $$ \int 2x\cos(x^2)\,dx, $$

the required antiderivative is $\sin x^2+C$. Why? Well, you just need to see it; which you would if you've mastered differentiation and the use of the chain rule (my apologies if this sounds condencending, I didn't mean for it to).

For the integral you give above $\int 2x\sqrt{1+x^2}\,dx$, you should recognize that ${- 2\over3}(1+x^2)^{3/2}+C$ would work.

With $u$ substitution you're thinking "the integral $$\int2x\sqrt{1+x^2}\,dx$$ is the same as the integral $$\int \sqrt u \,du$$ as long as $u$ is replaced by $1+x^2$ after evaluating the latter integral"

This $u$-substitution business is just a way of keeping track of things (which you'll eventually see after practicing with the method), especially when the integrand is the product of a composition of functions with a function that is ''almost'' the derivative of the inner function of the composition.

I hope this helps. I could elaborate on the last paragraph if you think that would help further.

David Mitra
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  • I don't understand the $2xcos(x^2)$ example, I don't know how to do products for antiderivatives so I wouldn't know how to do it but if I break it down (which I know I can't do) it would be $2, sin, 1/3x^3$ –  Nov 08 '11 at 19:19
  • You want an $\it antiderivative$ of $2x\cos (x^2)$; that is, a function whose derivative is $2x\cos (x^2)$. The antiderivative is $\sin x^2$ since ${d\over dx}\sin x^2=\cos( x^2)\cdot (x^2)'=2x\cos(x^2)$ – David Mitra Nov 08 '11 at 19:27
  • I don't understand how you know that, the derivative of $sinx^2$ is not something I would have to use something like the chain rule on? –  Nov 08 '11 at 19:36
  • $\sin (x^2)$ is a composition of functions ($sin x$ and $x^2$). You need the chain rule to differentiate it. Though I really don't like doing things this way: ''the derivative of the function "$\sin$ of something" is $\cos$ of the something times the derivative of the something. You do recognize that $\sin x^2$ is a composition ($\sin {\rm\ of\ } x^2$)? – David Mitra Nov 08 '11 at 19:43
  • So wouldnt it be like $(sinx)^2$ and the antiderivative would be $1/3 cosx^3$ ? –  Nov 08 '11 at 19:53
  • no, $\sin(x^2)$ and $(\sin x)^2$ are different (in the former, you square $x$ first). Also, the antiderivative of $(\sin x)^2$ is not $(1/3)(\sin x)^3$; but the antiderivative of $(\sin x)^2\cdot \cos x$ $\it is$ $(1/3)(\sin x)^3$. – David Mitra Nov 08 '11 at 19:59
  • So it is like sinu and u is $x^2$ so I have $cosx^2(2x)$? So the antiderivative would be (-cosx^2)(1/3x^3)$ ? –  Nov 08 '11 at 20:01
  • @Jordan: (i) You are composing "first $\sin x$, then the square". When composing functions, the order matters; here, you do "first the square, then the sine". (ii) The derivative of $\sin(x^2)$ is $2x\cos (x^2)$. The derivative of $(\sin x)^2$ is $2(\sin x)(\cos x)$. The antiderivative of $(\sin x)^2$ is not just $\frac{1}{3}(\cos x)^3$, because the derivative of the latter is not $(\sin x)^2$; it's $(\cos x)^2(-\sin x)$ by the Chain Rule. – Arturo Magidin Nov 08 '11 at 20:02
  • Yes, If I understand you correctly; except you start with $2x\cos x^2$ and replace with $\sin u$ (in the integral, the dx becomes du also). It seems we can't ''chat'' here, so this will be my last comment. – David Mitra Nov 08 '11 at 20:04
  • I don't understand this at all, I think I just have to wait until someone can explain it to me. –  Nov 08 '11 at 20:45
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I'll probably get stoned to death for such a clumsy answer, but here's how I do substitution problems:

Basically what you need to do is to get a just a single variable under the square root - that's the $u$ you are substituting. What you want to do is to find a substitution of a form that after derivation produces a differential that will "nullify" the $2x$ from $2x\sqrt{1 + x^2}$

So you go like this:

$$ \begin{align*} u =& 1 + x^2\\ du =& 2x \ dx\\ dx =&\frac{du}{2x} \end{align*} $$

So now when you subsitute:

$$ \begin{align*} \int \frac{2x}{2x}\sqrt{u}\ du\\ \int 1\sqrt{u}\ du\\ \int \sqrt{u}\ du \end{align*} $$

So the integral is: $$ \int \sqrt{u}\ du= \frac{2}{3}u^{\frac{3}{2}} = \frac{2}{3}(1+x^2)^{\frac{3}{2}} $$

Alexandar Živkovič
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