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Maths C4 - Integration... Help??? Watch

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    My teacher did teach us the reverse chain rule however he stressed very hard that you can only use it when function inside brackets is linear. Don't know if that's fully true but good so that it that rule doesn't even pass through my mind during c4 and I just use it when doing c3.
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    (Original post by black1blade)
    My teacher did teach us the reverse chain rule however he stressed very hard that you can only use it when function inside brackets is linear. Don't know if that's fully true
    Yes it's fully true.
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    (Original post by black1blade)
    My teacher did teach us the reverse chain rule however he stressed very hard that you can only use it when function inside brackets is linear. Don't know if that's fully true but good so that it that rule doesn't even pass through my mind during c4 and I just use it when doing c3.
    Your teacher is correct and taught the topic well. So many teachers don't which is why we get so many students on TSR who think that e.g.

    \displaystyle \int e^{x^2} \ dx = \frac{1}{2x} e^{x^2}+c
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    (Original post by notnek)
    Assuming you mean (f)

    Look at your formula book and see which of these terms you can integrate. Then let us know which terms you're still stuck on.
    Thank you! For some reason I forgot I could use the identity that  1 + cot^2 \theta \equiv cosec^2 \theta

    So I managed to get...

     = \int (cot^2 x - 2 cot x cosec x + cosec^2 x ) dx

     = \int (cosec^2 x - 1 - 2 cot x cosec x + cosec^2 x ) dx

     = -2cot x - x + 2cosec x + c

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    (Original post by Philip-flop)
    Thank you! For some reason I forgot I could use the identity that  1 + cot^2 \theta \equiv cosec^2 \theta

    So I managed to get...

     = \int (cot^2 x - 2 cot x cosec x + cosec^2 x ) dx

     = \int (cosec^2 x - 1 - 2 cot x cosec x + cosec^2 x ) dx

     = -2cot x - x + 2cosec x + c

    Yes, that is correct
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    Ok, so now I'm stuck on how to integrate these types...

     \int (3cos x sin^2 x) dx

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    (Original post by Philip-flop)
    Ok, so now I'm stuck on how to integrate these types...

     \int (3cos x sin^2 x) dx

    I tell right away it will be \sin^3(x)+c but whenever you see these types, just realise that cosine and sine are a product here, and they are both derivatives of each other (ignoring the sign change which is a constant mult.). So it would be a good idea to use substitution. Which one do you think would be a good one to sub in for?
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    (Original post by Philip-flop)
    Ok, so now I'm stuck on how to integrate these types...

     \int (3cos x sin^2 x) dx

    Any thoughts?

    If you have no idea then try differentiating \sin^3 x then try \sin^4 x and \sin^5 x. You should notice a pattern (you may know this pattern already).

    Getting used to these patterns will save you having to make a substitution in the exam (which may or may not be given to you). And I find that if you discover the pattern yourself as I've told you to above then it helps you remember it.
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    (Original post by RDKGames)
    I tell right away it will be \sin^3(x)+c but whenever you see these types, just realise that cosine and sine are a product here, and they are both derivatives of each other (ignoring the sign change which is a constant mult.). So it would be a good idea to use substitution. Which one do you think would be a good one to sub in for?
    How on earth did you know instantly? I'm really struggling with these types I'm not even sure I know how to use Integration by Substitution in this case

    (Original post by notnek)
    Any thoughts?

    If you have no idea then try differentiating \sin^3 x then try \sin^4 x and \sin^5 x. You should notice a pattern (you may know this pattern already).

    Getting used to these patterns will save you having to make a substitution in the exam (which may or may not be given to you). And I find that if you discover the pattern yourself as I've told you to above then it helps you remember it.
    Ok, so I can see (worked out) that...
     f(x) = sin^3 x  \Rightarrow f'(x) = 3 cosx sin^2 x

     f(x) = sin^4 x \Rightarrow f'(x) = 4 cosx sin^3 x

     f(x) = sin^5 x  \Rightarrow f'(x) = 5 cosx sin^4 x << I didn't check whether this one was correct I just assumed.
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    (Original post by Philip-flop)
    How on earth did you know instantly? I'm really struggling with these types I'm not even sure I know how to use Integration by Substitution in this case



    Ok, so I can see (worked out) that...
     \int sin^3 x = 3 cosx sin^2 x

     \int sin^4 x = 4 cosx sin^3 x

     \int sin^5 x = 5 cosx sin^4 x << I didn't check whether this one was correct I just assumed.
    Use the one which has the higher exponent on it, so sine has the higher exponent. Thus u=\sin(x) works.

    Also, those integral signs should be derivative signs
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    (Original post by Philip-flop)
    How on earth did you know instantly? I'm really struggling with these types I'm not even sure I know how to use Integration by Substitution in this case



    Ok, so I can see (worked out) that...
     \int sin^3 x = 3 cosx sin^2 x

     \int sin^4 x = 4 cosx sin^3 x

     \int sin^5 x = 5 cosx sin^4 x << I didn't check whether this one was correct I just assumed.
    Yes so this shows you if you need to integrate something of the form \cos x multiplied by a power of \sin x then you can "reverse the chain rule" by raising the power of the \sin x by 1, ignoring coefficients for now. Can you see this?

    Similarly you could integrate \sin x \cos^6 x by considering the derivative of \cos^7 x.

    It takes a lot of practice to get confident with this type of integration. Please keep asking if you need help.
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    (Original post by RDKGames)
    Use the one which has the higher exponent on it, so sine has the higher exponent. Thus u=\sin(x) works.

    Also, those integral signs should be derivative signs
    Yes I'm not sure why I used  \int instead of  f'(x) haha, can you tell my brain is fried?

    Awesome, I'll try using the substitution that  u = sin(x) now

    Thank you!

    (Original post by notnek)
    Yes so this shows you if you need to integrate something of the form \cos x multiplied by a power of \sin x then you can "reverse the chain rule" by raising the power of the \sin x by 1, ignoring coefficients for now. Can you see this?

    Similarly you could integrate \sin x \cos^6 x by considering the derivative of \cos^7 x.

    It takes a lot of practice to get confident with this type of integration. Please keep asking if you need help.
    Yes I'm starting to see it now. I never properly came across these types until now so I guess I need to carry on familiarising myself with them!

    Thank you so much!
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    (Original post by Philip-flop)
    Yes I'm not sure why I used  \int instead of  f'(x) haha, can you tell my brain is fried?

    Awesome, I'll try using the substitution that  u = sin(x) now

    Thank you!



    Yes I'm starting to see it now. I never properly came across these types until now so I guess I need to carry on familiarising myself with them!

    Thank you so much!
    By the way, this integral pattern isn't mentioned in the spec but is in the Edexcel textbook. And I've seen a few past paper questions that can be done almost instantly if you are confident with this integral form.

    What is mentioned on the spec are integrals like

    \displaystyle \int \frac{3x}{2x^2+4} \ dx

    Again you can use a substitution but the best method here is to recognise the pattern.
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    (Original post by Philip-flop)
    Yes I'm not sure why I used  \int instead of  f'(x) haha, can you tell my brain is fried?

    Awesome, I'll try using the substitution that  u = sin(x) now

    Thank you!

    Yes I'm starting to see it now. I never properly came across these types until now so I guess I need to carry on familiarising myself with them!

    Thank you so much!
    For fun you can try proving that \displaystyle \int (n+1)\cos(x)\sin^n(x) .dx = \sin^{n+1}(x)+c where n \geq 1 using substitution. I found it always nice to test my knowledge at A-Level by considering the general examples.

    Then switch 'em around and prove that \displaystyle \int (n+1)\sin(x)\cos^n(x).dx = -\cos^{n+1}(x)+c where n \geq 1 using an appropriate substitution

    Might help you become more used to the pattern here and these problems in general.

    Extension:
    Spoiler:
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    \displaystyle \int f'(x)[f(x)]^n .dx = \frac{[f(x)]^{n+1}}{n+1} +c, \quad n \geq 1



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    (Original post by RDKGames)
    ...
    Why n\geq 1?
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    (Original post by Zacken)
    Why n\geq 1?
    No particular reason other than it fits in with his example more.
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    (Original post by notnek)
    By the way, this integral pattern isn't mentioned in the spec but is in the Edexcel textbook. And I've seen a few past paper questions that can be done almost instantly if you are confident with this integral form.

    What is mentioned on the spec are integrals like

    \displaystyle \int \frac{3x}{2x^2+4} \ dx

    Again you can use a substitution but the best method here is to recognise the pattern.
    I'm not even sure I even know how to solve \displaystyle \int \frac{3x}{2x^2+4} \ dx using Integration by Substitution let alone by recognition

    Feel like giving up.

    I think I know how to integrate by using the fact that...  \int \frac{f'(x)}{f(x)} dx = ln |f(x)| + c

    (Original post by RDKGames)
    For fun you can try proving that \displaystyle \int (n+1)\cos(x)\sin^n(x) .dx = \sin^{n+1}(x)+c where n \geq 1 using substitution. I found it always nice to test my knowledge at A-Level by considering the general examples.

    Then switch 'em around and prove that \displaystyle \int (n+1)\sin(x)\cos^n(x).dx = -\cos^{n+1}(x)+c where n \geq 1 using an appropriate substitution

    Might help you become more used to the pattern here and these problems in general.

    Extension:
    Spoiler:
    Show








    \displaystyle \int f'(x)[f(x)]^n .dx = \frac{[f(x)]^{n+1}}{n+1} +c, \quad n \geq 1







    Yes I definitely have to give this a go!
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    (Original post by Philip-flop)
    I'm not even sure I even know how to solve \displaystyle \int \frac{3x}{2x^2+4} \ dx using Integration by Substitution let alone by recognition
    You really need to know how to do these.

    You need to spot that the numerator is a constant away from the derivative of 2x^2+4. And when you spot this pattern, the integral is always \ln (\text{the denominator})+c and then you adjust the constant in front of the log.

    You should know this since the derivative of \ln f(x) is \frac{f'(x)}{f(x)} so if you are integrating a fraction where the numerator is the derivative of the denominator then by reversing the differentiation you end up with \ln f(x).
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    (Original post by Philip-flop)
    I'm not even sure I even know how to solve \displaystyle \int \frac{3x}{2x^2+4} \ dx using Integration by Substitution let alone by recognition

    Feel like giving up.

    I think I know how to integrate by using the fact that...  \int \frac{f'(x)}{f(x)} dx = ln |f(x)| + c



    Yes I definitely have to give this a go!
    Go through the integration chapter in the textbook again if you're not feeling comfortable with substitutions or recognitions.

    The more you practice recognition questions... the easier they are to recognise.
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    (Original post by notnek)
    You really need to know how to do these.

    You need to spot that the numerator is a constant away from the derivative of 2x^2+4. And when you spot this pattern, the integral is always \ln (\text{the denominator})+c and then you adjust the constant in front of the log.

    You should know this since the derivative of \ln f(x) is \frac{f'(x)}{f(x)} so if you are integrating a fraction where the numerator is the derivative of the denominator then by reversing the differentiation you end up with \ln f(x).
    Yeah I know how to integrate by using the fact that...  \int \frac{f'(x)}{f(x)} dx = ln |f(x)| + c but I wasn't sure how I would do this by Substitution that's all
 
 
 
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