Turn on thread page Beta
    • Thread Starter
    Offline

    7
    ReputationRep:


    Alright so I'm confused on how to integrate this with recognition / applying chain rule in reverse.

    I know it can be solved with other methods eg: IBS, but would like to know how to do it, the other way.
    Offline

    15
    ReputationRep:
    Consider the derivative of  \sin(x^2) .
    Offline

    22
    ReputationRep:
    (Original post by SaadKaleem)


    Alright so I'm confused on how to integrate this with recognition / applying chain rule in reverse.

    I know it can be solved with other methods eg: IBS, but would like to know how to do it, the other way.
    The important thing to note here is that integration by substitution is exactly the same thing as recognition/applying the chain rule in reverse, just a slightly different way of looking at it.

    So here, you can see you have an integrand of the form \frac{d}{dx}(x^2) \cos (x^2) \, \mathrm{d}x. So you can immediately see that it'll integrate to something like \sin(x^2) (up to a constant factor blah blah).

    So what you do is \frac{d}{dx}(\sin (x^2)) = 2x \cos (x^2)

    Your integrand is \frac{1}{3}x\cos (x^2) so you want to turn the 2 into a 1/3.

    The way to do this is: \frac{1}{3 \times 2}\frac{d}{dx}(\sin(x^2)) = \frac{1}{3}x\cos (x^2)

    So integrating \frac{1}{3}x\cos (x^2) gives you \frac{1}{6}\sin(x^2) by recognition.

    With enough practice, you'll be able to skip the step of writing down \frac{d}{dx}(sin (x^2)) = 2x\cos (x^2) and be able to jump straight to the answer, I will stress that this comes with practice, so for now - stick with writing the derivative.
    Offline

    15
    ReputationRep:
    (Original post by SaadKaleem)


    Alright so I'm confused on how to integrate this with recognition / applying chain rule in reverse.

    I know it can be solved with other methods eg: IBS, but would like to know how to do it, the other way.
    Alright so I usually have a hard time explaining integration by recognition but here goes.

    First of all, take some sort of a "guess". Smart guess of course.

    With this integral I would say y=sin(x^{2})
    As the integral of cosx is sinx
    This line is basically saying my guess is that this will integrate to sin(x^{2})

    Now, once you've made this guess, you differentiate it to see if you get what you're trying to integrate.

    y=sin(x^{2})
    \dfrac{dy}{dx} = 2xcos(x^{2})

    You can see this is very close to the main integral, but its 6 times too big.

    So from that you will know that the main question integrates to \dfrac{1}{6}sin(x^{2})

    I hate explaining integration by recognition, I find it hard to explain :laugh:
    Though Zacken has beat me to it anyway
    • Thread Starter
    Offline

    7
    ReputationRep:
    (Original post by B_9710)
    -----.
    (Original post by Zacken)
    -----
    (Original post by edothero)
    -----
    Wow thank you all for the fast responses, yeah that was a bit dumb moment for me, Should've considered sin(x^2)... lol

    Anyways, I've understood it.
    Offline

    22
    ReputationRep:
    (Original post by SaadKaleem)
    Wow thank you all for the fast responses, yeah that was a bit dumb moment for me, Should've considered sin(x^2)... lol

    Anyways, I've understood it.
    Hmm... 2x\cos(x^2) \neq \frac{1}{6} \cdot 2x \cos(x^2).
    • Thread Starter
    Offline

    7
    ReputationRep:
    (Original post by Zacken)
    Hmm... 2x\cos(x^2) \neq \frac{1}{6} \cdot 2x \cos(x^2).
    Damn, I've been making that mistake, I guess.

    But can't it be used as a check that multiplying by 1/6 would equal the dy/dx value to the original integrand? So, we'd now also need to multiply by 1/6 to the value we've considered?
    Offline

    22
    ReputationRep:
    (Original post by SaadKaleem)
    Damn, I've been making that mistake, I guess.

    But can't it be used as a check that multiplying by 1/6 would equal the dy/dx value to the original integrand? So, we'd now also need to multiply by 1/6 to the value we've considered?
    I'm just saying that your notation is off.

    Don't write \frac{dy}{dx} = 2x\cos (x^2) = \frac{1}{6}\cdot 2x \cos(x^2).

    Write: \frac{dy}{dx} = 2x\cos(x^2) so \frac{1}{6} \frac{dy}{dx} = \frac{1}{6}\cdot 2x \cos(x^2) = \frac{1}{3}x\cos(x^2).

    Remember, equal signs mean that's what's on the left is exactly what's on the right.
    • Thread Starter
    Offline

    7
    ReputationRep:
    (Original post by Zacken)
    I'm just saying that your notation is off.

    Don't write \frac{dy}{dx} = 2x\cos (x^2) = \frac{1}{6}\cdot 2x \cos(x^2).

    Write: \frac{dy}{dx} = 2x\cos(x^2) so \frac{1}{6} \frac{dy}{dx} = \frac{1}{6}\cdot 2x \cos(x^2) = \frac{1}{3}x\cos(x^2).

    Remember, equal signs mean that's what's on the left is exactly what's on the right.
    Ah, I see. Will definitely be careful on that.

    Thank you as always Zacken!
    Offline

    22
    ReputationRep:
    (Original post by SaadKaleem)
    Ah, I see. Will definitely be careful on that.

    Thank you as always Zacken!
    No problem.
    Offline

    3
    ReputationRep:
    (Original post by Zacken)
    No problem.
    Hi, I was wondering if you can also do this with integration by parts? I tried to do it but I got no where
    Offline

    22
    ReputationRep:
    (Original post by SirRaza97)
    Hi, I was wondering if you can also do this with integration by parts? I tried to do it but I got no where
    Nopes, not possible.
    Offline

    3
    ReputationRep:
    (Original post by Zacken)
    Nopes, not possible.
    Why not? Is it not 2 functions?
    Offline

    22
    ReputationRep:
    (Original post by SirRaza97)
    Why not? Is it not 2 functions?
    So? It's not just because it's two functions that it means that it's integrable by parts.
    Offline

    3
    ReputationRep:
    (Original post by Zacken)
    So? It's not just because it's two functions that it means that it's integrable by parts.
    Then what am I missing? Because in a C4 exam I would've thought "Oh that's integration by parts". I am obviously not understanding something.
    Offline

    22
    ReputationRep:
    (Original post by SirRaza97)
    Then what am I missing? Because in a C4 exam I would've thought "Oh that's integration by parts". I am obviously not understanding something.
    The fact that the Fresnel integrals aren't really available to A-Level students... In a C4 exam, you'd be given the sub of u=x^2 or meant to do it by recognition as above.
    Offline

    3
    ReputationRep:
    (Original post by Zacken)
    The fact that the Fresnel integrals aren't really available to A-Level students... In a C4 exam, you'd be given the sub of u=x^2 or meant to do it by recognition as above.
    Okay... your explanations are very bland. But thanks anyways.
    Offline

    22
    ReputationRep:
    (Original post by SirRaza97)
    Okay... your explanations are very bland. But thanks anyways.
    No problem.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: May 16, 2016
Poll
Favourite type of bread
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

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

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