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Integration by inspection

Hi, I understand how to do integration by inspection but I just got confused by one example so I would appreciate it if somebody could help me.

∫cos3xdx \int cos^3xdx

I'll show what the solution says.

∫cos3xdx=∫cosxcos2xdx=∫d(sinx)(1−sin2x)dx=sinx−sin3x3 \int cos^3xdx = \int cosxcos^2xdx = \int d(sinx)(1-sin^2x)dx = sinx-\frac{sin^3x}{3}

I just don't understand how ∫d(sinx)(1−sin2x)dx=sinx−sin3x3 \int d(sinx)(1-sin^2x)dx = sinx-\frac{sin^3x}{3}

It feels like there is something extremely simple that I'm missing so any pointers on what happened during that step would be nice. Thanks in advance.
Reply 1
Avatar for TeeEm
TeeEm
19
Original post by Aiden223
Hi, I understand how to do integration by inspection but I just got confused by one example so I would appreciate it if somebody could help me.

∫cos3xdx \int cos^3xdx

I'll show what the solution says.

∫cos3xdx=∫cosxcos2xdx=∫d(sinx)(1−sin2x)dx=sinx−sin3x3 \int cos^3xdx = \int cosxcos^2xdx = \int d(sinx)(1-sin^2x)dx = sinx-\frac{sin^3x}{3}

I just don't understand how ∫d(sinx)(1−sin2x)dx=sinx−sin3x3 \int d(sinx)(1-sin^2x)dx = sinx-\frac{sin^3x}{3}

It feels like there is something extremely simple that I'm missing so any pointers on what happened during that step would be nice. Thanks in advance.


sine differentiates to cosine
Reply 2
Avatar for Notnek
Notnek
21
Original post by Aiden223
Hi, I understand how to do integration by inspection but I just got confused by one example so I would appreciate it if somebody could help me.

∫cos3xdx \int cos^3xdx

I'll show what the solution says.

∫cos3xdx=∫cosxcos2xdx=∫d(sinx)(1−sin2x)dx=sinx−sin3x3 \int cos^3xdx = \int cosxcos^2xdx = \int d(sinx)(1-sin^2x)dx = sinx-\frac{sin^3x}{3}

I just don't understand how ∫d(sinx)(1−sin2x)dx=sinx−sin3x3 \int d(sinx)(1-sin^2x)dx = sinx-\frac{sin^3x}{3}

It feels like there is something extremely simple that I'm missing so any pointers on what happened during that step would be nice. Thanks in advance.

Do you understand what d(sinx)d(sinx) means? That reffers to the derivative of sinx. Maybe it will be clearer for you if it's kept as cosx:

∫cos3xdx=∫cosxcos2xdx=∫cosx(1−sin2x)dx \displaystyle \int cos^3xdx = \int cosxcos^2xdx = \int cosx(1-sin^2x)dx

=∫cosx dx−∫cosxsin⁡2x dx\displaystyle = \int cosx \ dx-\int cosx\sin^2x \ dx

The first integral gives sinx so it's just the second integral to worry about:

∫cosxsin⁡2x dx\displaystyle \int cosx\sin^2x \ dx

This is the one that can be done by inspection. Can you see how? Or is this the part where you're stuck?
(edited 9 years ago)
Reply 3
Avatar for Aiden223
Aiden223
OP
7
Original post by notnek
Do you understand what d(sinx)d(sinx) means? That reffers to the derivate of sinx. Maybe it will be clearer for you if it's kept as cosx:

∫cos3xdx=∫cosxcos2xdx=∫cosx(1−sin2x)dx \displaystyle \int cos^3xdx = \int cosxcos^2xdx = \int cosx(1-sin^2x)dx

=∫cosx dx−∫cosxsin⁡2x dx\displaystyle = \int cosx \ dx-\int cosx\sin^2x \ dx

The first integral gives sinx so it's just the second integral to worry about:

∫cosxsin⁡2x dx\displaystyle \int cosx\sin^2x \ dx

This is the one that can be done by inspection. Can you see how? Or is this the part where you're stuck?


Ooh I see, yeah the d(sinx) d(sinx) bit confused me. Thank you!
Reply 4
Avatar for Aiden223
Aiden223
OP
7
Original post by notnek
Do you understand what d(sinx)d(sinx) means? That reffers to the derivative of sinx. Maybe it will be clearer for you if it's kept as cosx:

∫cos3xdx=∫cosxcos2xdx=∫cosx(1−sin2x)dx \displaystyle \int cos^3xdx = \int cosxcos^2xdx = \int cosx(1-sin^2x)dx

=∫cosx dx−∫cosxsin⁡2x dx\displaystyle = \int cosx \ dx-\int cosx\sin^2x \ dx

The first integral gives sinx so it's just the second integral to worry about:

∫cosxsin⁡2x dx\displaystyle \int cosx\sin^2x \ dx

This is the one that can be done by inspection. Can you see how? Or is this the part where you're stuck?


Actually I think it is the ∫cosxsin2dx \int cosxsin^2dx bit that confuses me, I don't understand how to do that by inspection?
Reply 5
Avatar for Notnek
Notnek
21
Original post by Aiden223
Actually I think it is the ∫cosxsin2dx \int cosxsin^2dx bit that confuses me, I don't understand how to do that by inspection?

Think about what the derivative is of e.g. cos3xcos^3 x or sin⁡3x\sin^3 x or cos⁡4x\cos^4 x or any sin⁡nx\sin^n x or cos⁡nx\cos^n x. What connect them?


Then the next question is: what differentiates to give cosxsin2xcosxsin^2 x. You should be able to get the general form and then adjust with a constant.

Over time you will recognise these forms and be able to see what they will integrate to.
(edited 9 years ago)
Original post by Aiden223
Actually I think it is the ∫cosxsin2dx \int cosxsin^2dx bit that confuses me, I don't understand how to do that by inspection?


ddxsin⁡2x=... \dfrac {d}{dx} \sin^2 x = ...
Reply 7
Avatar for ztibor
ztibor
10
Original post by Aiden223
Hi, I understand how to do integration by inspection but I just got confused by one example so I would appreciate it if somebody could help me.

∫cos3xdx \int cos^3xdx

I'll show what the solution says.

∫cos3xdx=∫cosxcos2xdx=∫d(sinx)(1−sin2x)dx=sinx−sin3x3 \int cos^3xdx = \int cosxcos^2xdx = \int d(sinx)(1-sin^2x)dx = sinx-\frac{sin^3x}{3}

I just don't understand how ∫d(sinx)(1−sin2x)dx=sinx−sin3x3 \int d(sinx)(1-sin^2x)dx = sinx-\frac{sin^3x}{3}



This integral is not correct in this form

As d(sin⁡x)dx=cos⁡x→d(sin⁡x)=cos⁡x⋅dx\frac{d(\sin x)}{dx}=\cos x \rightarrow d(\sin x)=\cos x\cdot dx

the d(sin⁡x) d(\sin x) substitutes cos⁡x⋅dx\cos x\cdot dx product in
the original integral

so the integral in correct form will be

∫(1−sin⁡2x)d(sin⁡x)\int (1-\sin^2 x) d(\sin x)

that is you have to integrate wrt. sinx
Original post by Aiden223
Hi, I understand how to do integration by inspection but I just got confused by one example so I would appreciate it if somebody could help me.

∫cos3xdx \int cos^3xdx

I'll show what the solution says.

∫cos3xdx=∫cosxcos2xdx=∫d(sinx)(1−sin2x)dx=sinx−sin3x3 \int cos^3xdx = \int cosxcos^2xdx = \int d(sinx)(1-sin^2x)dx = sinx-\frac{sin^3x}{3}

I just don't understand how ∫d(sinx)(1−sin2x)dx=sinx−sin3x3 \int d(sinx)(1-sin^2x)dx = sinx-\frac{sin^3x}{3} To take a slightly different approach: if we get rid of the dx (*), we have

∫(1−sin⁡2x) d(sin⁡x)\int (1-\sin^2x)\, d(\sin x)

Now, everywhere we have sin x, we want to treat it as a single object. To make that more obvious, let's replace it with U:

∫1−U2 dU\int 1 - U^2 \, dU

of course we can integrate this: ∫1−U2 dU=U−13U3+C\int 1 - U^2 \, dU = U - \frac{1}{3} U^3 + C

now, replace the U's with sin x's again: ∫(1−sin⁡2x) d(sin⁡x)=sin⁡x−13sin⁡3x+C\int (1-\sin^2x)\, d(\sin x) = \sin x - \frac{1}{3}\sin^3 x + C.

With practice, you can do the "treat sin x as a single object" thing without needing to replace it with a single letter, at which point you can go straight from ∫(1−sin⁡2x) d(sin⁡x)\int (1-\sin^2x)\, d(\sin x) to ∫(1−sin⁡2x) d(sin⁡x)=sin⁡x−13sin⁡3x+C\int (1-\sin^2x)\, d(\sin x) = \sin x - \frac{1}{3}\sin^3 x + C in one step.

(*) The d(sin x) notation is somewhat "dodgy", but if you are going to use it, it seems to me you must get rid of the dx (both because it "works" better but also because what you're "really" doing here is rewriting dsin⁡xdx dx\dfrac{d \sin x}{dx} \, dx and the "dx" terms cancel (**)).

(**) Could the analysts stop screaming please - this is a methods question!
Note that in DFranklin's "replace sin(x) with U" step, he's *not* integrating by substitution, but simply formally replacing the term sin(x) with U.
Reply 10
Avatar for ztibor
ztibor
10
Original post by Smaug123
Note that in DFranklin's "replace sin(x) with U" step, he's *not* integrating by substitution, but simply formally replacing the term sin(x) with U.


You can consider that it is a simple substitution , I think.

replacing sinx with U and d(sinx) with dU.

With a substitution you have to change the variables/functions
with another variables/function and the dx differential with the proper differential
as integrating factor

If sinx=U then cosx dx =dU

or if there is no cosx as factor, then

x=arc sin U -> dx=1/\sqrt(1-U^2) dU
Original post by Smaug123
Note that in DFranklin's "replace sin(x) with U" step, he's *not* integrating by substitution, but simply formally replacing the term sin(x) with U.



Original post by ztibor
You can consider that it is a simple substitution , I think.
Yes, I agree that it's substitution. It's just a scenario where the substitution is simple enough that with practice we we can skip most of the steps.
For that matter, I've always disliked the phrase "integration by inspection", because one man's inspection might be another man's page of working, particularly if one person spends a lot of time dealing with certain integrals. Someone who often has to take moments of probability distributions might write:

∫x5e−x dx=−(x5+5x4+20x3+60x2+120x+120)e−x\int x^5 e^{-x}\,dx = -(x^5+5x^4+20x^3+60x^2+120x+120)e^{-x} "by inspection" but there's actually a *lot* of steps going on there (even if it's easy once you get the pattern).

For that matter, ∫f(x) dx\int f(x)\,dx is solved instantly by the "substitution" u(x)=∫0xf(t) dtu(x) = \int_0^x f(t)\, dt, so if the integral is actually solvable, there's always *some* substitution that solves it "by inspection".
Original post by DFranklin
For that matter, I've always disliked the phrase "integration by inspection", because one man's inspection might be another man's page of working, particularly if one person spends a lot of time dealing with certain integrals.

What they don't teach at A-level, I believe, is the fact that "inspection" means "I spotted by magic that the answer is made substantially simpler by <this>, and I will now verify the magic step properly". As far as I remember, it's often simply taught as a synonym for "reverse chain rule".
Reply 14
Avatar for davros
davros
16
Original post by DFranklin
For that matter, I've always disliked the phrase "integration by inspection", because one man's inspection might be another man's page of working,



Original post by Smaug123
What they don't teach at A-level, I believe, is the fact that "inspection" means "I spotted by magic that the answer is made substantially simpler by <this>, and I will now verify the magic step properly". As far as I remember, it's often simply taught as a synonym for "reverse chain rule".


I'll throw in my twopence worth here - mainly because there hasn't been a more appropriate place to put this before, even though I've been thinking about this for a good few years!

I personally dislike both phrases "integration by inspection" and "reverse chain rule". And I'm pretty sure that neither existed in the (admittedly formal) textbooks I used to teach myself calculus many years ago.

"Integration by inspection" to me gives the false impression that there's some "magic box" operating, and that people with special skills of perception can "see" the answer, whereas weaker students cannot! I would always prefer (and always have done myself) to see students work through a substitution systematically (unless there's a very, very obvious differential hiding in there.), just to see that the student actually understands what's going on "under the hood". This is, perhaps, and old-fashioned view, and I note that the "young turks" on TSR seem to think that it's a weakness if students can't write down an integral "by inspection".

"Reverse chain rule" is even more of an anathema to me though. One of the weaknesses I see in modern students is an over-reliance on having some named rule for everything in existence, rather than trying to work things out from first principles, and one of the big weaknesses in the modern A level presentation of calculus is the failure to distinguish between differentiation, where a finite number of rules will always work (at least for any finite combination of functions encountered at A level), and integration, where it's easy to conjure up examples of simple composite functions that just can't be integrated! From what I've seen on TSR in recent years, use of the phrase "reverse chain rule" preys on both weaknesses and leads many students to believe that with any sort of product or ratio or (especially) composition of functions they can just start integrating one part and dividing by another, Then they get confused when someone points out that integration doesn't work like this, because they want to fall back on the fact that something like (ax + b)^n seems to work when integrated using their technique!

It's very sad, I think, that students are no longer given an integral and expected to work out their own best approach to it using substitution and/or IBP as we used to have to do - every integration question now seems to be accompanied by the phrase "by using the substitution..." or "by integrating by parts..." to spell out how to do it!


That's my rant over for today :biggrin:
Original post by davros
I'll throw in my twopence worth here - mainly because there hasn't been a more appropriate place to put this before, even though I've been thinking about this for a good few years!Think you must have missed an opportunity then, because I'm sure I've posted my side of this rant before!

I personally dislike both phrases "integration by inspection" and "reverse chain rule". And I'm pretty sure that neither existed in the (admittedly formal) textbooks I used to teach myself calculus many years ago.
I think I might have seen integration by inspection mentioned by Siklos at some point. Never heard of "reverse chain rule" before TSR.

"Integration by inspection" to me gives the false impression that there's some "magic box" operating, and that people with special skills of perception can "see" the answer, whereas weaker students cannot! I would always prefer (and always have done myself) to see students work through a substitution systematically (unless there's a very, very obvious differential hiding in there.), just to see that the student actually understands what's going on "under the hood". This is, perhaps, and old-fashioned view, and I note that the "young turks" on TSR seem to think that it's a weakness if students can't write down an integral "by inspection".
I have pretty much the same view. I must say, when I was "in training" for S-levels/Tripos, I was probably a fair bit quicker at these kinds of integrals. But I'd still say that it's making mistakes that really costs me time with tough integrals, not being slow and methodical with the substitutions.

"Reverse chain rule" is even more of an anathema to me though.
Don't get me started...

One of the weaknesses I see in modern students is an over-reliance on having some named rule for everything in existence, rather than trying to work things out from first principles, and one of the big weaknesses in the modern A level presentation of calculus is the failure to distinguish between differentiation, where a finite number of rules will always work (at least for any finite combination of functions encountered at A level), and integration, where it's easy to conjure up examples of simple composite functions that just can't be integrated! From what I've seen on TSR in recent years, use of the phrase "reverse chain rule" preys on both weaknesses and leads many students to believe that with any sort of product or ratio or (especially) composition of functions they can just start integrating one part and dividing by another,
Well, I think the true "horror" of the "reverse chain rule" is that most people seem to have it wrong.

(As you know), it's true that ∫g′(x)f′(g(x)) dx=f(g(x))\int g'(x) f'(g(x)) \,dx = f(g(x)), and this seems to be the standard "correct" way of stating the inverse chain rule.

But somehow, most people seem to think you can divide both sides by g'(x) to get:

]∫f′(g(x)) dx=1g′(x)f(g(x))]\int f'(g(x)) \,dx = \frac{1}{g'(x)} f(g(x))

I'd love to know where this came from, because it's so common I feel someone must be teaching it this way.

Incidentally, googling "reverse chain rule" doesn't leave me feeling that this is a particularly well founded technique. Looks like a bit of a "quirk" that has somehow become very popular with A-level / TSR students.

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