Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    2
    ReputationRep:
    When we substitute x = tan (theta) to find definite integration, why we assume theta to be an acute angle?
    Offline

    12
    ReputationRep:
    Would need context but the only reason I can think of off the top of my head is that tan is positive when acute and negative 90<theta<180
    Offline

    22
    ReputationRep:
    (Original post by mystreet091234)
    When we substitute x = tan (theta) to find definite integration, why we assume theta to be an acute angle?
    We don't? We simple restrict x = \tan \theta to be piecewise injective over our integration domain. This means that we need 0 \leq \theta \leq \pi (or any other interval of length \pi, with the obvious exception of \theta = \frac{\pi}{2} in which case we get an improper integral. You might want to give an example or such so that we can clarify.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Zacken)
    We don't? We simple restrict x = \tan \theta to be piecewise injective over our integration domain. This means that we need 0 \leq \theta \leq \pi (or any other interval of length \pi, with the obvious exception of \theta = \frac{\pi}{2} in which case we get an improper integral. You might want to give an example or such so that we can clarify.
    How do we know if the upper limit is not pi-pi/6 for obtuse angle?
    Q:Name:  question.jpg
Views: 99
Size:  7.7 KB
    A:Attachment 513987513989
    Attached Images
     
    Offline

    22
    ReputationRep:
    (Original post by mystreet091234)
    How do we know if the upper limit is not pi-pi/6 for obtuse angle?
    Because, like I said - we need x \mapsto \sin x to be an injective mapping, i.e: \arcsin \frac{1}{2} can only have one value and we pick the principal value \frac{\pi}{6}.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Zacken)
    Because, like I said - we need x \mapsto \sin x to be an injective mapping, i.e: \arcsin \frac{1}{2} can only have one value and we pick the principal value \frac{\pi}{6}.
    I see thanks!
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Zacken)
    We don't? We simple restrict x = \tan \theta to be piecewise injective over our integration domain. This means that we need 0 \leq \theta \leq \pi (or any other interval of length \pi, with the obvious exception of \theta = \frac{\pi}{2} in which case we get an improper integral. You might want to give an example or such so that we can clarify.
    Sorry I'm still confused. Why does it have to be injective?
    Name:  injective.jpg
Views: 107
Size:  35.1 KB
    For example, why isn't there a plus/minus sign near the u in step (iii) ?

    Also for finding Cartesian equation of a curve from its parametric equations x=sin t and y=sin 2t, are the two equations also injective mapping?Name:  image.jpg
Views: 81
Size:  515.6 KB

    But for another question, it seems the marking scheme isnt treating the equation as injective...when should an equation be treated as injective?
    Name:  image.jpg
Views: 83
Size:  232.9 KBAttachment 514541514543

    Sorry for my dumbness
    Zacken aymanzayedmannan
    Attached Images
     
    Offline

    3
    ReputationRep:
    (Original post by mystreet091234)
    Sorry for my dumbness
    You're not being dumb - it's actually a really good thing that you're attempting to understand why we're using a one-to-one function rather than just learning rote methods.

    I'll let Zacken handle your confusion about injectivity because he'll have a much more intuitive answer.


    (Original post by mystreet091234)
    But for another question, it seems the marking scheme isnt treating the equation as injective...when should an equation be treated as injective?
    Name:  image.jpg
Views: 83
Size:  232.9 KB
    However, for this question I don't really see anything about restricting domains as such, this seems like straightforward differentiation.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by aymanzayedmannan)
    You're not being dumb - it's actually a really good thing that you're attempting to understand why we're using a one-to-one function rather than just learning rote methods.

    I'll let Zacken handle your confusion about injectivity because he'll have a much more intuitive answer.




    However, for this question I don't really see anything about restricting domains as such, this seems like straightforward differentiation.
    Thanks for replying so quickly!!
    I'm waiting for Zacken's reply too
    You two are the besttt

    So for that question when we don't have to restrict domain, one-to-one function isn't needed?

    When do we need to restrict domain? for integration and parametric equation?
    Offline

    22
    ReputationRep:
    (Original post by mystreet091234)
    Sorry I'm still confused. Why does it have to be injective?
    It doesn't really need to be injective, it just needs to be piece wise injective. So for example, I can take u^2 = x^2 - 4 piecewise on (-\infty, 0] and [, \infty) where it is injective on these two different intervals separately. This is a bit hard to wrap your head around, I know - so I'm going to say that at this level, you can treat mostly everything as "just taking the positive root" - so if you have \cos^2 t = 1-\sin^2 t, don't worry about it - at A-Level you will only just need to take the positive root and write down \cos t = \sqrt{1-\sin^2 t}. I wish I had a better explanation but I don't have quite enough time to type one out right now. Don't worry about injectivity, it seems that I have just confused you by using the word.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Zacken)
    It doesn't really need to be injective, it just needs to be piece wise injective. So for example, I can take u^2 = x^2 - 4 piecewise on (-\infty, 0] and [, \infty) where it is injective on these two different intervals separately. This is a bit hard to wrap your head around, I know - so I'm going to say that at this level, you can treat mostly everything as "just taking the positive root" - so if you have \cos^2 t = 1-\sin^2 t, don't worry about it - at A-Level you will only just need to take the positive root and write down \cos t = \sqrt{1-\sin^2 t}. I wish I had a better explanation but I don't have quite enough time to type one out right now. Don't worry about injectivity, it seems that I have just confused you by using the word.
    Okay so for sub-fuctions of the piecewise function just take positive roots so this also explains my parametric equation problem!!!

    THANKS SO MUCH
    Offline

    3
    ReputationRep:
    (Original post by mystreet091234)
    Thanks for replying so quickly!!
    I'm waiting for Zacken's reply too
    You two are the besttt

    So for that question when we don't have to restrict domain, one-to-one function isn't needed?

    When do we need to restrict domain? for integration and parametric equation?
    One way to avoid thinking about it is by taking a sub of u = \sqrt{x^{2} -4}. We know that squaring both sides only yields the positive value of the function. Honestly, don't worry about restriction of domains because you'll only ever need the positive root in A-level maths and further maths.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by aymanzayedmannan)
    One way to avoid thinking about it is by taking a sub of u = \sqrt{x^{2} -4}. We know that squaring both sides only yields the positive value of the function. Honestly, don't worry about restriction of domains because you'll only ever need the positive root in A-level maths and further maths.
    Got it!!
    Thank you guys
    Offline

    22
    ReputationRep:
    (Original post by mystreet091234)
    Okay so for sub-fuctions of the piecewise function just take positive roots so this also explains my parametric equation problem!!!

    THANKS SO MUCH
    Here's a post I've made about injectivity and substitution in the past: http://www.thestudentroom.co.uk/show...9#post62689379 - you might find it vaguely illuminating.
 
 
 
  • 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.

  • Poll
    Has a teacher ever helped you cheat?
    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
  • 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

    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.