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**confused94****Member**- Registered: 2006-08-26
- Posts: 7

How do I integrate

(4x)

---------------

sqrt((5x^2)-1)

confused

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**Zhylliolom****Real Member**- Registered: 2005-09-05
- Posts: 412

Have you learned the technique of Integration by Substitution? Here's the basic idea of it:

Let's look at our integral:

In its current form, you probably don't recognize it as something you can integrate. but if it were in the form

you could solve it in a heartbeat. The goal of Integration by Substitution is to get the integral into a simple form like the one above, so you can evaluate the integral. How do we go about this? Well, I'm not sure how well I can explain this, and it may seem like I am making so leaps in logic, but this is just a thing you need to practice, and then it will become natural and you'll be able to know the correct substitution to make nearly all of the time.

Ok, now if we were to substitute some u in for x in a given integral ∫f(x) dx, by the chain rule the integral would turn into ∫f(u) du. This probably just looks confusing, doing the example will make it become clearer. So let's look at our integral here and find a suitable expression to set as u. The best way to go about this is to think of what integrals we can easily solve that look like the given integral and then what substitution x = u would be able to change the integral into that form. Looking at the integral, we may see the form

which we know we can solve easily. It turns out that we can get this form if we let u = 5x² - 1. We choose this substitution so that we can get only one term under the square root. Almost always with Integration by Substitution we wish to get only one term in places like (u)[sup]n[/sup], cos(u), e[sup]u[/sup], etc. Anyway, if we let u = 5x² - 1, then du = 10x dx (do you get what I did here?). Now we plug in u for 5x² - 1 and du for 10x dx(we multiplied 4x by 5/2 in the second step to get 4x to be 10x and to balance out that 5/2 we multiplied the entire integral by 2/5):

Now we just substitute 5x² - 1 for u to get our answer in terms of x:

I hope this was clear.

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**confused94****Member**- Registered: 2006-08-26
- Posts: 7

Thankyou, that was beautifully written. I actually learned alot from that. But what happened to the dx bit? It seemed to disappear. Shouldn't it have become 2/5 ∫ (u)^-1/2 du dx?

Sorry, confused again.

I understood your method completely though.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Remeber that du = 10x dx, so the dx is still in there, it's just hidden under the name of du.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**Zhylliolom****Real Member**- Registered: 2005-09-05
- Posts: 412

Yes, Ricky is right. dx is part of du, so it "disappears" when we make the substitution du = 10x dx.

It's funny how every student of the Calculus I have taught is always uncomfortable with the "disappearing dx" at first. It's just such a widespread thing that everyone seems to feel.

Thanks for calling it "beautifully written". I'm glad my work for you is appreciated .

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**confused94****Member**- Registered: 2006-08-26
- Posts: 7

You're a teacher? I should have known, there's difference between saying and teaching, it is a skill.

What I meant.. I understood du replaces 10x, since they are both equal, hence, du/√u

.....but why isn't there a dx at the end. Or alternatively, since dx = du/10x, a du/10x on the end? Ahhh...

Just realised...you did it slightly different, see: http://img242.imageshack.us/my.php?image=4atb6.png

You'd still end up with the same answer.

Those pesky du's and dx's.....They never explain what it means, they just tell you it. lol

Kind regards

*Last edited by confused94 (2006-08-27 07:48:28)*

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**confused94****Member**- Registered: 2006-08-26
- Posts: 7

Another question if I may, why do you multiply the whole integral by 2/5?

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**Zhylliolom****Real Member**- Registered: 2005-09-05
- Posts: 412

Because to get 4x to be 10x, so we can substitute 10x dx = du, we need to multiply 4x by 5/2 ((5/2)*4x = 10x). But this gives us a different integral (instead of ∫f(x) dx we now have ∫(5/2)f(x) dx). So to keep the integral the same, we multiply it by the reciprocal of 5/2, which is 2/5. This will work out since (2/5)∫(5/2)f(x) dx = ∫f(x) dx, so it is indeed the same integral. Is that clear?

Yes, the image you posted with the substitution dx = du/10x works as well. It is just a matter of preference. If you understand that method better, then perhaps you should use it instead.

I am actually a student, not a teacher. I call myself a teacher to sound cooler, and perhaps I am just an unofficial teacher. I do teach other math and physics if they let me, I just don't get a salary for it .

I think I may post a thread in the Exercises section about Integration by Substitution later tonight. Keep your eyes open for it, and when it shows up, try to do the exercises to see if you get the concept of Integration by Substitution. For now, I need to go do homework . I'll be back later.

Edit: I just realized that you said "why isn't there a dx on the end" for 10x = du. There is supposed to be: du = 10x dx. I'm not sure if I made a typo somewhere and forgot the dx or what. Just in case you're wondering, we get du = 10x dx from

*Last edited by Zhylliolom (2006-08-27 08:57:24)*

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**Zhylliolom****Real Member**- Registered: 2005-09-05
- Posts: 412

Also, note the following:

If you were to make the substitution dx = du/10x, you would have

since 4x/10x = 2/5. So from that, you have another way of seeing how the 2/5 showed up.

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**confused94****Member**- Registered: 2006-08-26
- Posts: 7

Now I understand. In order for the integral to remain unchanged, you must multiply the by the reciprocal 2/5 * 5/2 = 1. Therefore it is unchanged.

So if you were to multiply the numerator by 2, you would have to multiply the whole integral by 1/2.

How much is a 1/2 pound of your brain? I'd like to buy, if possible (you only use 10% after all!).

I'll have a go when I see it.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

How much is a 1/2 pound of your brain? I'd like to buy, if possible (you only use 10% after all!).

Yea, sign me up as well. But on a serious note, the whole %10 thing is one big myth that was spread by the media. If you think you only use 10% of your brain, which part could you do without?

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**Zhylliolom****Real Member**- Registered: 2005-09-05
- Posts: 412

Here are some exercises(I recommend that you don't try the double starred ones until after Calculus III(I don't think most standard curriculums even teach the proper way to do the double starred ones, I just put them there for people like Ricky to pull their hair out over)):

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**Zhylliolom****Real Member**- Registered: 2005-09-05
- Posts: 412

Hey confused94, just making sure you don't forget about the Integration by Substitution thread I made the other day in response to this post. Hopefully you can go to this thread and benefit from the practice on the problems.

Here is the link again:

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