You are Here: Home >< Maths

# What's the point of integration by inspection? Watch

1. What's the point of integration by inspection?
Firstly, i don't understand it. What does d/dx mean?

Secondly, isn't it easier to use integration by substitution? Is it for things like (u)^-1 which cant be done that way?

The explanation in the textbook is utterly ****.
2. d/dx then something after it means the derivative of it, e.g d/dx(x^2) = 2x.

It is usually faster than making a substitution
3. (Original post by Jacke02)
What's the point of integration by inspection?
Firstly, i don't understand it. What does d/dx mean?

Secondly, isn't it easier to use integration by substitution? Is it for things like (u)^-1 which cant be done that way?

The explanation in the textbook is utterly ****.
besides from being much quicker in some cases than substitution, you can also inspection to integrate things like tan and cot rather than using substitution
4. So you do not spend forever looking for suitable substitutions, i.e.

What is the integral of

?
5. because integration by substitution (sometimes) is an invalid argument for a lot functions that are to be integrated

for example if you try to integrate x^2 with the substitution u=x^2, it is not a valid argument, doing it by inspection and understanding the reverse-differentiation is a valid argument, faster, and is what separates you from a robot

and heres another example of why using integration by sub can result in disaster

try to integrate e^x^2 * cosx with the limits of (x, 0, 1), you'll see it cannot be done by substitution, or at all, really, but by INSPECTION we can use the trapezium rule on it, or explain how its an odd function, meaning it'll equal 0 every time, thus its a better argument to use inspection here

d/dx is confusing in that form --> d (a function here)/dx is much better, it means you are entering a function, such as d(2x)/dx which is then just 2, to be differentiated with respect to x, implying x is not a constant, this is important when it comes to understanding partial derivatives

edit: text books at a level and at the start of university tend to be dreadful, its a game of trial and error, picking authors that produce learning books that are suited to your style, try to look at some youtube videos on maths to help your learning on integration
6. (Original post by DeanK22)
So you do not spend forever looking for suitable substitutions, i.e.

What is the integral of

?
I would do that in my head with a substitution though.

To be honest, I'd never heard of integration by inspection up until a few weeks ago. I'm still not really sure what it is. Is it just a different way of phrasing the fundamental theorem of calculus?
7. (Original post by htn)
for example if you try to integrate x^2 with the substitution u=x^2, it is not a valid argument, doing it by inspection and understanding the reverse-differentiation is a valid argument, faster, and is what separates you from a robot
You'd have to be a little careful with the region you're integrating over, but it would be valid in some cases. I'm not convinced it would be a useful substitution though.

try to integrate e^x^2 * cosx with the limits of (x, 0, 1), you'll see it cannot be done by substitution, or at all, really, but by INSPECTION we can use the trapezium rule on it, or explain how its an odd function, meaning it'll equal 0 every time, thus its a better argument to use inspection here
That integral is not 0. The nth trapezium rule approximation is merely an approximation. You may be able to take limits but I can't remember precisely what that would involve in this case, and I'm not sure it would always be legitimate because of the difficulties in measure theory that limits bring about.
8. (Original post by DeanK22)
What is the integral of

?
9. (Original post by IrrationalNumber)
I would do that in my head with a substitution though.

To be honest, I'd never heard of integration by inspection up until a few weeks ago. I'm still not really sure what it is. Is it just a different way of phrasing the fundamental theorem of calculus?
It is an informal statement of the FTC it is not really something one hears of, simply a name or phrase certain people use to say they are using the FTC.
10. (Original post by Pheylan)
???
11. (Original post by Pheylan)
Don't know where you got that from the integral for this is ln(sinx+cosx) (by inspection)
12. (Original post by anshul95)
Don't know where you got that from the integral for this is ln(sinx+cosx) (by inspection)
Think (hope) he's taking the piss.
13. (Original post by anshul95)
Don't know where you got that from the integral for this is ln(sinx+cosx) (by inspection)

dean asked for the integral of , which is , no?
14. (Original post by Pheylan)

dean asked for the integral of , which is , no?
you've integrated twice - don't know why you have done that
15. (Original post by anshul95)
you've integrated twice - don't know why you have done that
whoosh

(the integral of an integral...)
16. (Original post by anshul95)
you've integrated twice - don't know why you have done that
but i just explained why i've done that
17. oh, I see, slight anti climax....
18. (Original post by Pheylan)
but i just explained why i've done that
no - dean asked by the first bit just look at what you have written again - and if you still don't believe differentiate ln(sin x+ cos x) - you should get
(cos x - sin x)/(cos x + sin x), which is the function dean wanted to integrate.
19. (Original post by anshul95)
no - dean asked by the first bit just look at what you have written again - and if you still don't believe differentiate ln(sin x+ cos x) - you should get
(cos x - sin x)/(cos x + sin x), which is the function dean wanted to integrate.
never mind.
20. Another way to look at it is "advanced guessing".

Imagine having a function 1/sqrt(1-x)

Rather than making the substitution and changing the variables again, you could just take a guess with the integral being sqrt(1-x). You differentiate mentally to check, and you find that it's missing a minus sign. So put it in.

The reason why some people find this method difficult is because it's something that gets easy only with experience. Recognize these patterns, like the numerator being the derivative of the denominator, the numerator being a constant and the denominator being the square root of some constant plus/minus x, etc and you will find that you can intuitively guess the integral. And to show that, you rewrite the integrand as the derivative of the integral (which you have guessed), and they 'cancel' out.

Think of it like in factorisation. Don't tell me that you are still drawing the 2x2 box to carry out quadratic factoring...

And d/dx is an operator, to find the first derivative of an expression with respect to x.

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: January 12, 2011
Today on TSR

### Oxford grad sues for £1m

Claims damages because he didn't get a first

### 'No evidence of shots fired' at Oxford Circus

Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

## Groups associated with this forum:

View associated groups
Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.