How Does U-Substitution Simplify Integration Problems?

[QuarkCharmer]

Homework Statement

Homework Equations

The Attempt at a Solution

I don't understand what exactly is going on here. They let $u=(1+x^{2})$, so that leaves them with this:
$$\int \frac{x}{u^{2}}dx$$

The derivative of $(1+x^{2})$ is simply $2x$. And so:
$$\frac{du}{dx} = 2x \rightarrow du = 2xdx \rightarrow dx=\frac{du}{2x}$$

So now, substituting in my new dx, I get:

$$\int \frac{x}{u^{2}2x}du$$

So, is that x simply canceling out here? Is that the idea?
Which leaves me with:
$$\int \frac{u^{-2}}{2}du$$
$$-\frac{1}{2u}$$
Re-substituting u I get:
$$-\frac{1}{2(x^{2}+1)} + C$$

With that being said, how do you know that the x will cancel? How are you even supposed to know that this approach will work? Is there some sort of proof to this idea, my book does not have it.

 

[rock.freak667]

So, is that x simply canceling out here? Is that the idea?

Yes that is the basic idea of it.

With that being said, how do you know that the x will cancel? How are you even supposed to know that this approach will work? Is there some sort of proof to this idea, my book does not have it.

We know it will work since d/dx(x²+1) = 2x, meaning that in the integrand the x in the numerator will cancel out.

 

[gb7nash]

So, is that x simply canceling out here?

That's pretty much the idea. Another way to look at it is:

$u=(1+x^{2})$ , $du= 2x dx$ , so $\frac{1}{2}du=xdx$.

Now substituting in, you have $\int \frac{1}{u^{2}} \frac{1}{2}du$ = ...

 

[micromass]

Homework Statement

Homework Equations

The Attempt at a Solution

I don't understand what exactly is going on here. They let $u=(1+x^{2})$, so that leaves them with this:
$$\int \frac{x}{u^{2}}dx$$

The derivative of $(1+x^{2})$ is simply $2x$. And so:
$$\frac{du}{dx} = 2x \rightarrow du = 2xdx \rightarrow dx=\frac{du}{2x}$$

So now, substituting in my new dx, I get:

$$\int \frac{x}{u^{2}2x}du$$

So, is that x simply canceling out here? Is that the idea?
Which leaves me with:
$$\int \frac{u^{-2}}{2}du$$
$$-\frac{1}{2u}$$
Re-substituting u I get:
$$-\frac{1}{2(x^{2}+1)} + C$$

Yes, that's it. You know that $u=x^2+1$, so deriving both sides gives you $du=2xdx$, thus $dx=\frac{du}{2x}$. And then you can substitute it in the integral.

With that being said, how do you know that the x will cancel? How are you even supposed to know that this approach will work? Is there some sort of proof to this idea, my book does not have it.

Basically, you don't know if the approach will work, you'll only know it by trying. You'll need some experience with these kind of thingies. If you've solved lots of integrals then you know immediately a few substitutions that you could try. But you'll still need to try them to see if they will really work out.

The same thing happens with the integral

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

when confronted with a root, I always try the substitution $u=\sqrt{x^2+1}$ first. Sometimes it works, sometimes it doesn't. With this substitution, you get $u^2=x^2+1$ and by deriving both sides:

$$2udu=2xdx$$

thus $dx=\frac{udu}{x}$

and you see that the x will cancel again. Note, if there wasn't an x before the root, then it wouldn't have worked. And if there wasn't an x in the numerator of your integral, then it wouldn't have worked...

 

[QuarkCharmer]

gb7nash,

$du= 2x dx$ , so $\frac{1}{2}du=dx$
Where did the x go in this example?

$du=2xdx$

$\frac{1}{2}du=xdx$

$\frac{1}{2x}du=dx$ ?

 

[gb7nash]

gb7nash,

$du= 2x dx$ , so $\frac{1}{2}du=dx$
Where did the x go in this example?

Typo. Thanks.

 

[QuarkCharmer]

Ah okay. The way you phrased it, I thought you were doing something different than me.

Thanks for all of the help everyone.

 

[S_Happens]

Personally after finiding du I would solve for xdx instead of for dx in this case, since that is what you have in the integral. Obvioulsy you can take the couple extra steps to see that it cancels out, but I don't see the reason.

At the point where you have 1/2du = xdx, I would simply make the substitution back into the integral, rather than making the extra steps that give you the same result.

 

[QuarkCharmer]

Related to this question:

$$\int sec^{3}(x)tan(x) dx$$

$u=sec(x)$, and so, $\frac{du}{dx}=sec(x)tan(x)$, and $$dx=\frac{du}{sec(x)tan(x)}$$

$$\int u^{3}tan(x) dx$$
$$\int \frac{u^{3}tan(x)}{sec(x)tan(x)} du$$
$$\int \frac{u^{3}}{sec(x)} du$$

I have no idea what to do with this one? Can I put my u=sec back in and try again now?

Edit: Na, that just gets me back to where I started.

Oh wait, because u is equal to sec, can I just call that $\frac{u^{3}}{u}$ ?

 

[rock.freak667]

I have no idea what to do with this one? Can I put my u=sec back in and try again now?

Edit: Na, that just gets me back to where I started.

You'd get

$$\int \frac{u^3}{u} du = \int u^2 du$$

 

[QuarkCharmer]

Yeah! I see what you did there (above edit). Fantastic stuff. I haven't had to actually think about math for a while, this is easily the funnest section I have done to date.

 

[vela]

How are you even supposed to know that this approach will work?

One thing you should always check is if the integral is of or can be tweaked into the form
$$\int [f(x)]^n f'(x)\, dx$$
If it is, the substitution u=f(x) will work, and the answer will be
$$\int [f(x)]^n f'(x)\, dx = \left\{ \begin{array}{lc} \frac{[f(x)]^{n+1}}{n+1}+c & n\ne -1 \\ \\ \log \lvert f(x) \rvert+c & n=-1 \end{array} \right.$$
In your problem, you'd guess f(x)=x²+1 so that f'(x)=2x, and it indeed works out:
$$\int \frac{x}{(x^2+1)^2}\,dx = \frac{1}{2}\int (x^2+1)^{-2} (2x) \,dx$$
If you can differentiate in your head, you can often an the integral fits the pattern by inspection. It's a neat trick to know mostly so you can screw with your peers when you look at an integral and say, "Oh, the answer is obviously..." and then write the answer down without showing any work. They'll be in awe of your mathematical prowess.

 

[vela]

For your latest problem, you'd say
$$\int \sec^3 x\tan x \,dx = \int (\sec x)^2 (\sec x \tan x)\,dx = \cdots$$

 

[QuarkCharmer]

For your latest problem, you'd say
$$\int \sec^3 x\tan x \,dx = \int (\sec x)^2 (\sec x \tan x)\,dx = \cdots$$

I came up with:

$$\frac{sec^3(x)}{3}+C$$


@Vela,
That is basically looking to see if it is a backwards chain rule correct?

 

[QuarkCharmer]

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

I let $u=\sqrt(1-x)$, and $dx = \frac{2\sqrt(1-x)du}{-1}$

So I came to this guy here:

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

But now if I resubstitute the $\sqrt(1-x)$ back in for u to cancel, I am still stuck with that x, and du at the end.

 

[micromass]

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

I let $u=\sqrt(1-x)$, and $dx = \frac{2\sqrt(1-x)du}{-1}$

So I came to this guy here:

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

It's easier not to work with the square roots:

IF $u=\sqrt{1-x}$, then $u^2=1-x$, thus $2udu=-dx$.

Furthermore, $x=1-u^2$, thus $x^2=(1-u^2)^2$.

Now you can substitute everything in the integral.

 

[QuarkCharmer]

Oh, I didn't even think about squaring both sides of the u-sub equation!
Would that work with other things too? For instance, suppose I say that u=MNOP, and then in the integral I see that there is a M/Mu, Could I multiply both sides by M, to get Mu = M^2NOP, and sub it into cancel the M in the numerator and such? Basically, is any normal operation okay there?

I'm working on it!

 

[vela]

That is basically looking to see if it is a backwards chain rule correct?

Yup.

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

I let $u=\sqrt(1-x)$, and $dx = \frac{2\sqrt(1-x)du}{-1}$

Which, since $u=\sqrt{1-x}$, gives you $dx=-2u\,du$. Then you need to express x in terms of u to finish off the substitution, which should give you the same thing micromass got.

So I came to this guy here:

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

But now if I resubstitute the $\sqrt(1-x)$ back in for u to cancel, I am still stuck with that x, and du at the end.

 

[QuarkCharmer]

Okay, I am giving this problem a shot now:

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

$$\frac{du}{dx}=\frac{-1}{2\sqrt{1-x}}$$

Since $\sqrt{1-x} = u$, then

$$\frac{du}{dx}=\frac{-1}{2u}$$
$$dx = -2udu$$

and so here is the squiggly guy:

$$\int \frac{-2ux^{2}}{u}du$$

and since $x=1-u^{2}$

$x^{2}=1-2u^{2}+u^{4}$

and so, putting all that back into the integrand:

$$\int \frac{-2u(1-2u^{2}+u^{4})}{u}du$$

Pair of u's cancel out

$$\int -2(1-2u^{2}+u^{4})du$$

$$\int -2+4u^{2}-2u^{4}du$$

Which is something that I can integrate easily...

$$-2u+\frac{4u^{3}}{3}-\frac{4u^{5}}{5}$$

To which, back go my original u's to get...

$$-2(\sqrt{1-x})+\frac{4(\sqrt{1-x})^{3}}{3}-\frac{4(\sqrt{1-x})^{5}}{5} + C$$

Which is a nightmare to simplify, that can't be right?

 

[vela]

I'd leave it as it. You can differentiate it and see if you recover the integrand.

 

[QuarkCharmer]

I found a much, much, much easier way to do this while playing around a bit. I'm too tired to latex it, but this one is correct as far as I can tell. The other answer might be right too, I didn't finish simplifying it.

The rest of my courses problems have definite integrals, and we have not covered that yet at all.

Again, thanks for all the help!

 

[kmacinto]

Also try it by letting u=(the entire denominator). For me, it's easier (and cleaner) to sub for the square root and then not have to worry about it (the sqr. root) again until converting back from u to x... Either way will work tho...

 

[QuarkCharmer]

Also try it by letting u=(the entire denominator). For me, it's easier (and cleaner) to sub for the square root and then not have to worry about it (the sqr. root) again until converting back from u to x... Either way will work tho...

I did that 2 posts up, in this case it seemed to be a bit more work expanding the roots back into the equation at the end.

 

[Char. Limit]

The rest of my courses problems have definite integrals, and we have not covered that yet at all.

Again, thanks for all the help!

Definite integrals are surprisingly easy once you know indefinite integrals. If you have a function f(x) such that $\int f(x) dx = F(x) + C$, then the definite integral of f(x) from a to b will be:

$$\int_a^b f(x) dx = F(b) - F(a)$$

The theorem that proves this is the Funadmental Theorem of Calculus.

 

[QuarkCharmer]

I'm teaching myself that this afternoon actually. I just got to the part that claims:
$$\int_a^b \! f(x) \, dx = lim_{n\to\infty} \sum_{i=1}^{n} f(x_{i}) \Delta x$$

No word on the fundamental theorem of calculus yet. Still struggling with understanding how using mid/left/right points will yield the same limit.

Alternatively, how do I make the roman "d" as in "dx" in latex? \dif does not appear to work?

 

[Char. Limit]

Well, that's mainly because as the number of "rectangles" increases (or goes to infinity), the difference between the area of all those rectangles and the area under the function will tend to 0, regardless of whether you use a left, right, or a mid point.

At least I think so.

EDIT: I don't understand...? I just type in dx in the latex and it gives me what i want. Unless you mean $\operatorname{d}x$, which is given by \operatorname{d}x.

 

[QuarkCharmer]

I'm still reading through the examples that use n=10 and other smaller numbers to do the rieman sum by hand.

Char wrote:
$$\int_a^b f(x) dx = F(b) - F(a)$$

That is, F(b)-F(a), where F(x) = the indefinite integral of f(x)? I have a hard time grasping how the area under the curve relates to the antiderivative. It seems like that constant would leave a huge margin for error. Idk, I'll get to that part I guess.

\dif is a command for the differential operator. It is simply an upface d to
clarify that it is an operator. E.g.:
\dif x dx

from: http://www.tug.org/texlive/Contents/live/texmf-dist/doc/latex/commath/commath.pdf

 

[micromass]

Still struggling with understanding how using mid/left/right points will yield the same limit.

This is not obvious to prove in a rigorous way. The proof of this uses things like uniform continuity of the integral and Cauchy sequences. Too much for a first encounter. I suggest you just accept this and go on.

Alternatively, how do I make the roman "d" as in "dx" in latex? \diff does not appear to work?

I think the standard is to use \mathrm{d}x.

 

[micromass]

I'm still reading through the examples that use n=10 and other smaller numbers to do the rieman sum by hand.

Char wrote:
$$\int_a^b f(x) dx = F(b) - F(a)$$

That is, F(b)-F(a), where F(x) = the indefinite integral of f(x)? I have a hard time grasping how the area under the curve relates to the antiderivative. It seems like that constant would leave a huge margin for error. Idk, I'll get to that part I guess.

That the antiderivative relates to the area is one of the most beautiful and surprising results from mathematics. I can't give you an intuitive reason why this is the case, you just need to read the proof and convince yourself from that fact. It's quite a short proof if I remember well.

 

[QuarkCharmer]

I suggest you just accept this and go on.

I already have, I was just thinking that the book would explain it, at least the idea, but it looks like it's jumping right into some sample definite integral problems instead. Ah well. I don't even know what a cauchy is

I can't give you an intuitive reason why this is the case, you just need to read the proof and convince yourself from that fact. It's quite a short proof if I remember well.

I will do that.

 

[micromass]

I already have, I was just thinking that the book would explain it, at least the idea, but it looks like it's jumping right into some sample definite integral problems instead. Ah well. I don't even know what a cauchy is

There are two ways to learn calculus: intuitive and easy, or rigorous with proofs. You need to know it intuitively before you can do it rigorously. But that means that the books will leave out facts that you don't find obvious at first. You'll need a more formal book if you want to understand everything , but I wouldn't advise this when first confronted with definite integrals

 

[QuarkCharmer]

I'm reading from the Stewart book, and using the Spivak book as a reference, because I find it hard to understand sometimes.

Well, that's mainly because as the number of "rectangles" increases (or goes to infinity), the difference between the area of all those rectangles and the area under the function will tend to 0, regardless of whether you use a left, right, or a mid point.

At least I think so.

Now that I think about it, this makes sense, because as the number of "n" increases, the width of each "n" decreases, and since $x_i$ is always within $\Delta x$, the squeeze theorem would dictate that as $\Delta x$ approaches 0, so does $x_i$. Does that make sense?

 

[Char. Limit]

I'm reading from the Stewart book, and using the Spivak book as a reference, because I find it hard to understand sometimes.



Now that I think about it, this makes sense, because as the number of "n" increases, the width of each "n" decreases, and since $x_i$ is always within $\Delta x$, the squeeze theorem would dictate that as $\Delta x$ approaches 0, so does $x_i$. Does that make sense?

That makes sense to me, and is probably a good intuitive basis for this fact. Now, the Fundamental Theorem of Calculus, that doesn't really have much of an intuitive basis, unfortunately.

 

[vela]

That makes sense to me, and is probably a good intuitive basis for this fact. Now, the Fundamental Theorem of Calculus, that doesn't really have much of an intuitive basis, unfortunately.

It sort of does, actually, but I've never found it all that helpful. Say you have a function f(x). At x=x₀, it has a value f(x₀). You can approximate the value of the function at a nearby point x₁=x₀+Δx as
$$f(x_1)=f(x_0+\Delta x) \approx f(x_0) + f'(x_0) \Delta x$$
You can see this easily from the graph. The intuition is simply that f'(x₀) tells you which way the function is going at x=x₀. If you know where you are, i.e., f(x₀), and which way you're going, i.e., f'(x₀), you can figure out where you'll end up, i.e., f(x₁). If you rearrange it slightly, you already see an inkling of the fundamental theorem:
$$f(x_1) - f(x_0) \approx f'(x_0)\Delta x$$

Repeating the process, at the point x₂=x₁+Δx, you obtain
$$f(x_2)=f(x_1+\Delta x) \approx f(x_1) + f'(x_1) \Delta x \approx f(x_0) + f'(x_0) \Delta x + f'(x_1) \Delta x$$
and so on. Now let x₀=a, x_n=b, and Δx=(b-a)/n. Then x_i=x₀+i Δx, and
$$f(b) \approx f(a) + f'(a) \Delta x + f'(x_1) \Delta x + \cdots + f'(x_{n-1})\Delta x$$
or
$$f(b)-f(a) \approx f'(a) \Delta x + f'(x_1) \Delta x + \cdots + f'(x_{n-1})\Delta x$$
The righthand side is a Riemann sum, and in the limit Δx→0, you get
$$f(b)-f(a) = \int_a^b f'(x)\,dx$$

 

[QuarkCharmer]

I found a video of this rather strange looking man doing the proof:


Going to check to see if it's proven on ocw.mit.