What's the point of integration by inspection?

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 ****.

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Reply 1

username219321

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

Reply 2

anshul95

Original post by Jacke02

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

Reply 3

Oh I Really Don't Care

So you do not spend forever looking for suitable substitutions, i.e.

What is the integral of

\displaystyle \int \frac{cos(x) - sin(x)}{cos(x) + sin(x)} dx

Reply 4

htn

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

(edited 13 years ago)

Reply 5

IrrationalNumber

Original post by DeanK22

So you do not spend forever looking for suitable substitutions, i.e.

What is the integral of

\displaystyle \int \frac{cos(x) - sin(x)}{cos(x) + sin(x)} dx

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?

Reply 6

IrrationalNumber

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.

Reply 7

Pheylan

Original post by DeanK22

What is the integral of

\displaystyle \int \frac{cos(x) - sin(x)}{cos(x) + sin(x)} dx

\displaystyle ax+b+\int ln(sinx+cosx)\ dx

Reply 8

Oh I Really Don't Care

Original post by IrrationalNumber

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.

Reply 9

elixir

Original post by Pheylan

\displaystyle ax+b+\int ln(sinx+cosx)\ dx

???

Reply 10

anshul95

Original post by Pheylan

\displaystyle ax+b+\int ln(sinx+cosx)\ dx

Don't know where you got that from the integral for this is ln(sinx+cosx) (by inspection)

Reply 11

Dirac Delta Function

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.

Reply 12

Pheylan

Original post by anshul95

Don't know where you got that from the integral for this is ln(sinx+cosx) (by inspection)

\displaystyle\int\frac{cosx-sinx}{cosx+sinx}\ dx=ln(cosx+sinx)+a

dean asked for the integral of

\displaystyle\int\frac{cosx-sinx}{cosx+sinx}\ dx

, which is

\displaystyle\int\int\frac{cosx-sinx}{cosx+sinx}\ dx\ dx

, no?

Reply 13

anshul95

Original post by Pheylan

\displaystyle\int\frac{cosx-sinx}{cosx+sinx}\ dx=ln(cosx+sinx)+a

dean asked for the integral of

\displaystyle\int\frac{cosx-sinx}{cosx+sinx}\ dx

, which is

\displaystyle\int\int\frac{cosx-sinx}{cosx+sinx}\ dx\ dx

, no?

you've integrated twice - don't know why you have done that

Reply 14

around

Original post by anshul95

you've integrated twice - don't know why you have done that

whoosh

(the integral of an integral...)

Reply 15

Pheylan

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 :confused:

Reply 16

elixir

oh, I see, slight anti climax....

Reply 17

anshul95

Original post by Pheylan

but i just explained why i've done that :confused:

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.

Reply 18

Pheylan

Original post by anshul95

never mind.

Reply 19

valhalla92

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.