1. Why is it that you differentiate when you integrate a chain function? Isn't the function already differentiated to begin with? So why is it necessary to differentiate it again in order to find the integral? It doesn't make any sense to me.

2. What happens with the right-hand side of the function when you integrate it? For example, with Example 5, there is an (x^3) in the function, but in the course of your working it out, it suddenly disappears and I have no idea where it went. With Example 6, there's a (3x^2) which goes AWOL as well.

3. In the course of your working out of Example 7 you say "The question has xdx so we write". Umm... where exactly did you get xdx from? When I look at the question I can't see any xdx.

Sorry if I'm sounding annoying, I'm just really confused at the moment.

Hi Murray,
I'm really, really confused about your explanation of integration on this page:
<a href="/integration/2-indefinite-integral.php">2. Antiderivatives and The Indefinite Integral</a>
and I really need help understanding it.
1. Why is it that you differentiate when you integrate a chain function? Isn't the function already differentiated to begin with? So why is it necessary to differentiate it again in order to find the integral? It doesn't make any sense to me.
2. What happens with the right-hand side of the function when you integrate it? For example, with Example 5, there is an (x^3) in the function, but in the course of your working it out, it suddenly disappears and I have no idea where it went. With Example 6, there's a (3x^2) which goes AWOL as well.
3. In the course of your working out of Example 7 you say "The question has xdx so we write". Umm... where exactly did you get xdx from? When I look at the question I can't see any xdx.
Sorry if I'm sounding annoying, I'm just really confused at the moment.

Relevant page
<a href="/integration/2-indefinite-integral.php">2. Antiderivatives and The Indefinite Integral</a>
What I've done so far
Read it and re-read it, but cannot grasp it.

First, relax! And you are not being annoying - your questions (as usual) are perfectly reasonable.

I think this is a case where you missed out on one important bit, and the rest of the page made no sense. Not surprising! I hope the following helps.

The key statement for the part that makes no sense is:

This is different to the other exercises above! It is a function of a function situation, so we have to do the reverse of the Chain Rule, which we met in the section on differentiation.

The new rule we require is...
Power Formula for Integration

The integrals above this statement (Examples 1 to 4) are simple - you just integrate each part step-by-step.

But Example 5 (and all the ones following) cannot be done step by step, because they have got 2 parts to them [in Ex 5, the 2 parts are the `(2x^4 - 5)^4` part and the `x^3` part].

In Ex 5, we could multiply it all out, then integrate the result, but it would be long and tedious. In Examples 7 and 8, we don't even have the option of multiplying it out, since it involves square roots. We need a new method to solve this type.

So the trick is (and it is a trick we must use to turn something complicated into something easier that we can do) is to consider the Chain Rule and then reverse it to find the integral.

Reminder: When we differentiated something like this: `y = (x^4 + 7)^5` using Chain Rule, we put `u = x^4 + 7 `and this means `y = u^5` then proceeded:

`dy/dx`

`= (dy/(du))((du)/dx)`

`= (5u^4)(4x^3)`

Writing this replacing back the `u` gives us

`= 5(x^4 + 7)^4 (4x^3)`

`= 20x^3(x^4 + 7)^4`

Now, looking at this answer, it is similar to the question in Ex 5.

So to integrate our Ex 5 question, we need to "undo" the Chain Rule. So instead of `dy/(du)` in the Chain Rule (that's differentiating), I need to do u^(n+1)/(n+1) (that's integrating the "`u`" part).

But if that's all I do, the final answer will not be a correct integral. I need to do the "`du`" part properly. That's why you see a differentiation step in each example. The "`du`" part is the differential which is part of all integral problems. See

The "AWOL" parts are essential for doing the problem. If we can't get those parts to "disappear", then we can't do the problem. But they are not magically disappearing, the method used is forcing them out.

Think of it (a bit) like this. If I give you this question:

`2x/6 = 5`

One way to do it (and the best because it makes it easier) is to cancel down the fraction on the left, to give:

`x/3 = 5`

`x = 15`

Hey, where did the `2` and `6` go? They were simplified (by cancelling, so they are AWL - absent with leave) to make the problem easier. Similarly, this Power Formula is a method of simplifying the integral so that we can do it.

Hopefully that gives you enough background to have another go through it. I will revise the page when I get a chance to add some of this more detailed explanation.

Now, let me be honest here. I remember that (when I was a student) I never got this whole section either, until I checked my answers. It only became clear why it worked and why we needed to do each step once I did this.

So for example, once I differentiated the answer for Ex 7, `(2x^4 - 5)^7/56 + K` and saw that my answer was equal to the question (the integral that I needed to find) then I convinced myself that it worked, and more importantly, I got a better insight into why it worked and the importance of that (pesky) differential step in the middle.

Re Example 7: The fraction in brackets has an "x" on the top, correct?

I could have written the question as

`int (x dx) / sqrt(x^2 + 9)`

Or I could have written it as

`int 1 / sqrt(x^2 + 9) (x dx)`

All I've done is changed the position of the x, but not the value of the thing being integrated. Either way, there is an "`x dx`" in the question.

X

Hi Daniel
First, relax! And you are not being annoying - your questions (as usual) are perfectly reasonable.
I think this is a case where you missed out on one important bit, and the rest of the page made no sense. Not surprising! I hope the following helps.
The key statement for the part that makes no sense is:
<blockquote>This is different to the other exercises above! It is a function of a function situation, so we have to do the reverse of the Chain Rule, which we met in the section on differentiation.
The new rule we require is...
Power Formula for Integration</blockquote>
The integrals above this statement (Examples 1 to 4) are simple - you just integrate each part step-by-step.
But Example 5 (and all the ones following) cannot be done step by step, because they have got 2 parts to them [in Ex 5, the 2 parts are the `(2x^4 - 5)^4` part and the `x^3` part].
In Ex 5, we could multiply it all out, then integrate the result, but it would be long and tedious. In Examples 7 and 8, we don't even have the option of multiplying it out, since it involves square roots. We need a new method to solve this type.
So the trick is (and it is a trick we must use to turn something complicated into something easier that we can do) is to consider the Chain Rule and then reverse it to find the integral.
Reminder: When we differentiated something like this: `y = (x^4 + 7)^5` using Chain Rule, we put `u = x^4 + 7 `and this means `y = u^5` then proceeded:
`dy/dx`
`= (dy/(du))((du)/dx)`
`= (5u^4)(4x^3)`
Writing this replacing back the `u` gives us
`= 5(x^4 + 7)^4 (4x^3)`
`= 20x^3(x^4 + 7)^4`
Now, looking at this answer, it is similar to the question in Ex 5.
So to integrate our Ex 5 question, we need to "undo" the Chain Rule. So instead of `dy/(du)` in the Chain Rule (that's differentiating), I need to do u^(n+1)/(n+1) (that's integrating the "`u`" part).
But if that's all I do, the final answer will not be a correct integral. I need to do the "`du`" part properly. That's why you see a differentiation step in each example. The "`du`" part is the differential which is part of all integral problems. See
<a href="/integration/1-differential.php">1. The Differential</a>
The "AWOL" parts are essential for doing the problem. If we can't get those parts to "disappear", then we can't do the problem. But they are not magically disappearing, the method used is forcing them out.
Think of it (a bit) like this. If I give you this question:
`2x/6 = 5`
One way to do it (and the best because it makes it easier) is to cancel down the fraction on the left, to give:
`x/3 = 5`
`x = 15`
Hey, where did the `2` and `6` go? They were simplified (by cancelling, so they are AWL - absent with leave) to make the problem easier. Similarly, this Power Formula is a method of simplifying the integral so that we can do it.
Hopefully that gives you enough background to have another go through it. I will revise the page when I get a chance to add some of this more detailed explanation.
Now, let me be honest here. I remember that (when I was a student) I never got this whole section either, until I checked my answers. It only became clear why it worked and why we needed to do each step once I did this.
So for example, once I differentiated the answer for Ex 7, `(2x^4 - 5)^7/56 + K` and saw that my answer was equal to the question (the integral that I needed to find) then I convinced myself that it worked, and more importantly, I got a better insight into why it worked and the importance of that (pesky) differential step in the middle.
Re Example 7: The fraction in brackets has an "x" on the top, correct?
I could have written the question as
`int (x dx) / sqrt(x^2 + 9)`
Or I could have written it as
`int 1 / sqrt(x^2 + 9) (x dx)`
All I've done is changed the position of the x, but not the value of the thing being integrated. Either way, there is an "`x dx`" in the question.

you say that the problem `sqrt(x^2+1)` cannot be solved the regular way because "the question does not have "x dx"so we cannot solve it using any of the integration methods used above."

However, I scrolled up a bit to Question 1, which is `sqrt(2x-1)`. This question is virtually identical to the one you say cannot be solved in the regular way (in that neither of them have a `x dx`), yet you STILL used the regular method to integrate it. So I don't understand why you would be able use the regular method for one, and not the other. What gives? (Either you're contradicting yourself or I'm missing something).

X

Hi Murray,
Thanks for your quick reply. I think I'm starting to get the hang of it now, but there's still one nagging little thing that's perplexing me.
At the very end of this page:
<a href="/integration/4-definite-integral.php">4. The Definite Integral</a>
you say that the problem `sqrt(x^2+1)` cannot be solved the regular way because "the question does not have "x dx"so we cannot solve it using any of the integration methods used above."
However, I scrolled up a bit to Question 1, which is `sqrt(2x-1)`. This question is virtually identical to the one you say cannot be solved in the regular way (in that neither of them have a `x dx`), yet you STILL used the regular method to integrate it. So I don't understand why you would be able use the regular method for one, and not the other. What gives? (Either you're contradicting yourself or I'm missing something).

I've done a re-write of that page. I hope it is clearer now.

In question 1, `du = 2 dx`, and as there is no "`x`" outside the square root, I was able to integrate it (by substituting `dx` with `(du)/2`).

In that last one, the one that has a problem, inside the square root is `x^2 + 1` and if I try to do `du = 2x dx`, I find there is no `x` outside of the square root, so I am stumped.

Hope that helps.

X

Hi Daniel
I've done a re-write of that page. I hope it is clearer now.
In question 1, `du = 2 dx`, and as there is no "`x`" outside the square root, I was able to integrate it (by substituting `dx` with `(du)/2`).
In that last one, the one that has a problem, inside the square root is `x^2 + 1` and if I try to do `du = 2x dx`, I find there is no `x` outside of the square root, so I am stumped.
Hope that helps.