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Further maths integration help

Hi! Can someone help me with integrating root(1+x^2)? I don't understand why you need to substitute sinh x into the integral. This might be a stupid question but why can't we use reverse chain rule to solve this question? Thanks in advance!

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Reply 1
Original post by AM2929
Hi! Can someone help me with integrating root(1+x^2)? I don't understand why you need to substitute sinh x into the integral. This might be a stupid question but why can't we use reverse chain rule to solve this question? Thanks in advance!


Maybe a better question is "what would you do if you use reverse chain rule"? Now presumably you've learned integration by substitution (or u-sub. That's reverse chain rule btw), so what do you let u to be? Read the spoilers when you've done some scratch work:

Spoiler

(edited 10 months ago)
Reply 2
Original post by tonyiptony
Maybe a better question is "what would you do if you use reverse chain rule"? Now presumably you've learned integration by substitution (or u-sub. That's reverse chain rule btw), so what do you let u to be? Read the spoilers when you've done some scratch work:

Spoiler



But if I re-write the integral as (1+x^2)^1/2 and then use reverse chain rule, I get 1/3x(1+x^2)^3/2. Can you please explain to me why this doesn't work?
Reply 3
Original post by AM2929
But if I re-write the integral as (1+x^2)^1/2 and then use reverse chain rule, I get 1/3x(1+x^2)^3/2. Can you please explain to me why this doesn't work?


You dont get that integral? For the reverse chain rule the integrand needs to be a product of two exressions like
u' * f(u)
where u' represents du/dx. The integral would be F(u) where f=dF/du. You dont have a product? You could differentiate your answer and see what you get.

If you had
x * sqrt(1+x^2)
youd be good to go with the reverse chain rule or sub with u=1+x^2.
(edited 10 months ago)
Reply 4
Original post by mqb2766
You dont get that integral? For the reverse chain rule the integrand needs to be a product of two exressions like
u' * f(u)
where u' represents du/dx. The integral would be F(u) where f=dF/du. You dont have a product? You could differentiate your answer and see what you get.

If you had
x * sqrt(1+x^2)
youd be good to go with the reverse chain rule or sub with u=1+x^2.

Ahh ok, that makes sense now. Thank you so much! But how do we know that we have to substitute sinh x into this integral to solve it?
Reply 5
Original post by AM2929
Ahh ok, that makes sense now. Thank you so much! But how do we know that we have to substitute sinh x into this integral to solve it?


Simple answer is that it makes the integral simpler/its a standard substitution to try.
x = sinh(u)
as the 1+x^2 is a standard identity for cosh^2.

You could also do x=tan(u) that equates to sec^2 for similar reasons
(edited 10 months ago)
Reply 6
Original post by AM2929
Ahh ok, that makes sense now. Thank you so much! But how do we know that we have to substitute sinh x into this integral to solve it?


Usually when you see a quadratic under a square root, trig sub is the obvious thing to try.
In fact, any time you see a quadratic that you can't factorize, it's a call for trig sub.
Reply 7
Original post by mqb2766
Simple answer is that it makes the integral simpler/its a standard substitution to try.
x = sinh(u)
as the 1+x^2 is a standard identity for cosh^2.

You could also do x=tan(u) that equates to sec^2 for similar reasons

Got it, thanks:smile:
Reply 8
Original post by tonyiptony
Usually when you see a quadratic under a square root, trig sub is the obvious thing to try.
In fact, any time you see a quadratic that you can't factorize, it's a call for trig sub.

Ok, that makes sense now. Thank you!
Reply 9
Original post by AM2929
Got it, thanks:smile:


As a bit of background, you should almost "expect" to involve trig/hyperbolic functions in the integration problem if the integrand is of the form
sqrt(1+/-x^2)
as they correspond to the area between the x-axis and curves of the form
y^2 +/- x^2 = 1
so the usual circle/hyperbola where points can be expressed in terms of sin/cos and sinh/cosh.

The circle one so something like the integral of
sqrt(1-x^2)
from 0 to X can easily be shown with a sketch to be
triangle area + sector area
and the sector area must depend on arcsin(X) as it is based on the angle. So when youre performing integration like this you should expect (at least with one representation) to get trig/hyperbolic functions in there somehow. Newton also famously developed/used the binomial series associated with (1-x^2)^(1/2) to get a series for pi.
https://www.quantamagazine.org/how-isaac-newton-discovered-the-binomial-power-series-20220831/
https://www.youtube.com/watch?v=gMlf1ELvRzc&ab_channel=Veritasium

So a simple line like
x = sin(u)
or similar has a fair bit of history behind it.
(edited 10 months ago)
Reply 10
Original post by AM2929
But if I re-write the integral as (1+x^2)^1/2 and then use reverse chain rule, I get 1/3x(1+x^2)^3/2. Can you please explain to me why this doesn't work?


To add to the "you're misusing the reverse chain rule" comments:

You've started off by looking at (1+x^2)^1/2 and thinking "that looks like a function of 1+x^2".
For concreteness, define g(x) = 1+x^2, then what you want to integrate is g(x)^(1/2) (w.r.t. x).

Up to here, everything is fine. Where you went wrong is that you

(A) integrated g^(1/2) w.r.t. g

and then

(B) divided by dg/dx.

These two steps are in the wrong order. What would have been correct is to divide g^(1/2) by dg/dx and then integrate that. [Of course, just because it's mathematically correct doesn't mean you can integrate what you get after you divide].

Note that when I spell out what the steps in applying "the reverse chain rule" correctly, it's basically the same thing as integration by substitution, only you're not explicitly writing down all of the steps.

Once you understand this, you're a lot less likely to make mistakes by using the reverse chain rule incorrectly.
Reply 11
Original post by DFranklin
To add to the "you're misusing the reverse chain rule" comments:

You've started off by looking at (1+x^2)^1/2 and thinking "that looks like a function of 1+x^2".
For concreteness, define g(x) = 1+x^2, then what you want to integrate is g(x)^(1/2) (w.r.t. x).

Up to here, everything is fine. Where you went wrong is that you

(A) integrated g^(1/2) w.r.t. g

and then

(B) divided by dg/dx.

These two steps are in the wrong order. What would have been correct is to divide g^(1/2) by dg/dx and then integrate that. [Of course, just because it's mathematically correct doesn't mean you can integrate what you get after you divide].

Note that when I spell out what the steps in applying "the reverse chain rule" correctly, it's basically the same thing as integration by substitution, only you're not explicitly writing down all of the steps.

Once you understand this, you're a lot less likely to make mistakes by using the reverse chain rule incorrectly.

I'm sorry, but could you please explain again what you meant by integrating in the wrong order? I didn't get that part. Thanks!
Reply 12
Original post by AM2929
I'm sorry, but could you please explain again what you meant by integrating in the wrong order? I didn't get that part. Thanks!

I restrained myself from having a rant on here earlier because others were helping, but you really must avoid using the "reverse chain rule" - it's not a "rule" at all but a trick that only works if your integrand happens to be in a specific convenient form - which it usually won't be!

What DFranklin is basically saying is that if you have an integral like g1/2dx\int g^{1/2} dx you can't do (1/(dg/dx)) ×g1/2dg \times \int g^{1/2} dg - that isn't valid. If you apply integration by substitution - which is a universal rule for integration - you will see that you should be doing this:

(1/(dg/dx))g1/2dg\int (1/(dg/dx)) g^{1/2} dg.

Depending on the substitution, this may or may not be easier than the original integral :smile:

May I ask why you think "reverse chain rule" is a concept - is your teacher telling you to do this, or have you seen it referenced somewhere?
(edited 10 months ago)
Reply 13
Original post by AM2929
I'm sorry, but could you please explain again what you meant by integrating in the wrong order? I didn't get that part. Thanks!


DFranklin
What would have been correct is to divide g^(1/2) by dg/dx and then integrate that.

E.g. Suppose we wanted to integrate x^6 and we decided to think of it as (x^2)^3 [obviously this isn't the best approach for integrating x^6 but I wanted an example where the calculations are obvious].

So we've got g(x)=x^2, and we're integrating g(x)^3 wrt x.

The correct method is to now divide g^3 by dg/dx *before* trying to integrate wrt to g. Wecan write dg/dx as 2sqrt(g) so we get (1/2)g^(5/2), whose integral is (1/2)(2/7)g^(7/2) which equals x^7/7, the correct answer.

If instead you integrated g^3 wrt g (getting g^4/4 = x^8/4) *before* dividing by dg/dx = 2x, you'd get x^7/8 which is wrong.
Reply 14
Original post by davros
May I ask why you think "reverse chain rule" is a concept - is your teacher telling you to do this, or have you seen it referenced somewhere?

From my experience over the years, the "reverse chain rule" confusion seems to have arisen directly due to the official Edexcel A Level textbook which is used by the majority of A Level students in the UK. The topics below are taught in this order:



The order, terminology and lack of proper explanations used in these topics by Edexcel has lead to mass confusion for the last 10+ years and I wonder if Edexcel are even aware of this.

Of course teachers should be making sure that their students understand these topics fully but you we can't blame them for following the order of the textbook.
Reply 15
Original post by DFranklin
E.g. Suppose we wanted to integrate x^6 and we decided to think of it as (x^2)^3 [obviously this isn't the best approach for integrating x^6 but I wanted an example where the calculations are obvious].

So we've got g(x)=x^2, and we're integrating g(x)^3 wrt x.

The correct method is to now divide g^3 by dg/dx *before* trying to integrate wrt to g. Wecan write dg/dx as 2sqrt(g) so we get (1/2)g^(5/2), whose integral is (1/2)(2/7)g^(7/2) which equals x^7/7, the correct answer.

If instead you integrated g^3 wrt g (getting g^4/4 = x^8/4) *before* dividing by dg/dx = 2x, you'd get x^7/8 which is wrong.

Thank you for the clarification! I understand what I was doing wrong now.
Reply 16
Original post by davros
I restrained myself from having a rant on here earlier because others were helping, but you really must avoid using the "reverse chain rule" - it's not a "rule" at all but a trick that only works if your integrand happens to be in a specific convenient form - which it usually won't be!

What DFranklin is basically saying is that if you have an integral like g1/2dx\int g^{1/2} dx you can't do (1/(dg/dx)) ×g1/2dg \times \int g^{1/2} dg - that isn't valid. If you apply integration by substitution - which is a universal rule for integration - you will see that you should be doing this:

(1/(dg/dx))g1/2dg\int (1/(dg/dx)) g^{1/2} dg.

Depending on the substitution, this may or may not be easier than the original integral :smile:

May I ask why you think "reverse chain rule" is a concept - is your teacher telling you to do this, or have you seen it referenced somewhere?

Thanks, I will use the substitution method to solve these types of questions. My teacher taught us the reverse chain rule concept but I didn't fully understand when to use it. Sorry for the confusion.
Reply 17
Original post by Notnek
From my experience over the years, the "reverse chain rule" confusion seems to have arisen directly due to the official Edexcel A Level textbook which is used by the majority of A Level students in the UK. The topics below are taught in this order:



The order, terminology and lack of proper explanations used in these topics by Edexcel has lead to mass confusion for the last 10+ years and I wonder if Edexcel are even aware of this.

So, what happened to 11.3?

I'm kind of curious about what they actually teach in 11.2 and 11.4. I think if I *had* to follow those items, I'd literally just teach:

11.2 If you know f(x)dx=F(x)\int f(x)\,dx = F(x), then f(ax+b)dx=1aF(ax+b)\int f(ax+b)\,dx = \frac{1}{a} F(ax+b)
11.4 f(x)f(x)dx=ln(f(x))\int \frac{f'(x)}{f(x)}\,dx = \ln(f(x)) (plus I'd have previously taught that kf(x)dx=kf(x)dx\int k f(x)\,dx = k \int f(x)\,dx).

I'd try really hard not to mention "reverse chain rule". (I'd prove them directly by differentiation if required).

[Anyhow, I know I'm preaching to the converted...]
(edited 10 months ago)
Reply 18
Original post by DFranklin
So, what happened to 11.3?

I'm kind of curious about what they actually teach in 11.2 and 11.4. I think if I *had* to follow those items, I'd literally just teach:

11.2 If you know f(x)dx=F(x)\int f(x)\,dx = F(x), then f(ax+b)dx=1aF(ax+b)\int f(ax+b)\,dx = \frac{1}{a} F(ax+b)
11.4 f(x)f(x)dx=ln(f(x))\int \frac{f'(x)}{f(x)}\,dx = \ln(f(x)) (plus I'd have previously taught that kf(x)dx=kf(x)dx\int k f(x)\,dx = k \int f(x)\,dx).

I'd try really hard not to mention "reverse chain rule". (I'd prove them directly by differentiation if required).

[Anyhow, I know I'm preaching to the converted...]

11.3 is "Using trigonometric identities". I just didn't include it because it wasn't relevant to reverse chain rule.
Reply 19
Original post by DFranklin
So, what happened to 11.3?



[Monty Python]Rule 11.3: There is no rule 11.3[/Monty Python]

I do wonder how people managed to "do calculus" for 300+ years without any mention of "the reverse chain rule" appearing in textbooks :smile:

I think those "ifs" in 11.2 and 11.4 are doing a lot of heavy lifting - they don't seem to have penetrated the consciousness of the students judging by the numerous failed attempts to apply the "reverse chain rule" on TSR these days! Of course it's hard to tell how much is poor teaching, and how much is coming from self-studying - TSR posters aren't exactly a representative group in any sense.

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