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    (Original post by notnek)
    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
    Actually makes total sense when you think about it. I was trying to integrate e^(x^2) for like 30 minutes before looking it up and finding out it's impossible. Thing is that if you try do a reverse chain rule where the function inside the bracket is not linear, you're going to divide by another function of x. When you come to differentiate to check if you've got the same answer, you'll have another function to deal with that wouldn't have been there otherwise so not equal. I explained that really poorly but yeah.
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    (Original post by Philip-flop)
    Ok so all of the integrals can be worked out without the need of having to use Substitution? So Integration by Parts is still in there etc?

    I ask because I'm thinking of using it on...

     \int 2x(x^2-3)^4 dx

    You're not too confident with the pattern recognition / inspection methods of integration so you're going to find a lot of the integrals on that sheet very challenging. But actually these recognition/inspection methods aren't required that much in C4 exams. So don't make this the focus of your revision - keep doing as many past papers as you can.

    If you can spot the pattern then that's the main thing since once you've spotted it you can either reverse the differentiation (the quickest method) or you can make a substitution for the function whose derivative you can also see in the expression.

    E.g. \displaystyle \int 3x(4x^2+3)^5 \ dx

    You need to notice that 3x is a constant away from the derivative of 4x^2+3 so that means you make a substitution for 4x^2+3. This works because in the substitution method you find the derivative of 4x^2+3 and this ends up "cancelling" with the 3x.

    Or the quickest method is to consider the derivative of (4x^2+3)^6 but this takes a lot of practice.
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    (Original post by RDKGames)
    Substitute u=\sin(3x-2)

    Or recognise that it's almost in the form f'(x)[f(x)]^n
    Oh yes I'm still having trouble using f'(x)[f(x)]^n with these types e.g.  f(x) = sin^3 x \Rightarrow f'(x) = 3 cosx sin^2 x

    So I managed to work out...
     \int cos(3x-2)sin^2(3x-2) dx

     = \frac{1}{3} \int 3cos(3x-2)sin^2(3x-2) dx

     = \frac{1}{9}sin^3 (3x-2) + c


    Gonna attempt this by substitution now I will not be defeated!

    Edit: RDKGames Ok, maybe I actually am still confused with the example above! How is  \int cos x sin^2 x classified as a f'(x)[f(x)]^n type if the derivative of  sin^2 x is  2cos x sinx ??

    Wait... I think I see it now, it's cos  \int cos x sin^2 x can be re-written as  \int cos x (sinx)^2 so that fits the f'(x)[f(x)]^n criteria, right?
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    (Original post by Philip-flop)
    Oh yes I'm still having trouble using f'(x)[f(x)]^n with these types e.g.  f(x) = sin^3 x \Rightarrow f'(x) = 3 cosx sin^2 x

    So I managed to work out...
     \int cos(3x-2)sin^2(3x-2) dx

     = \frac{1}{3} \int 3cos(3x-2)sin^2(3x-2) dx

     = \frac{1}{9}sin^3 (3x-2) + c


    Gonna attempt this by substitution now I will not be defeated!

    Edit: RDKGames Ok, maybe I actually am still confused with the example above! How is  \int cos x sin^2 x classified as a f'(x)[f(x)]^n type if the derivative of  sin^2 x is  2cos x sinx ??
    You pick \sin(x) not \sin^2(x)
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    (Original post by RDKGames)
    You pick \sin(x) not \sin^2(x)
    Wait... I think I see it now, it's cos  \int cos x sin^2 x can be re-written as  \int cos x (sinx)^2 so that fits the f'(x)[f(x)]^n criteria, right?
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    (Original post by Philip-flop)
    Wait... I think I see it now, it's cos  \int cos x sin^2 x can be re-written as  \int cos x (sinx)^2 so that fits the f'(x)[f(x)]^n criteria, right?
    Yes.
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    (Original post by RDKGames)
    Yes.
    I feel so stupid! I spent so long trying to figure this out over something that is actually so so so simple!! Thanks for clearing this up!


    (Original post by notnek)
    You're not too confident with the pattern recognition / inspection methods of integration so you're going to find a lot of the integrals on that sheet very challenging. But actually these recognition/inspection methods aren't required that much in C4 exams. So don't make this the focus of your revision - keep doing as many past papers as you can.

    If you can spot the pattern then that's the main thing since once you've spotted it you can either reverse the differentiation (the quickest method) or you can make a substitution for the function whose derivative you can also see in the expression.

    E.g. \displaystyle \int 3x(4x^2+3)^5 \ dx

    You need to notice that 3x is a constant away from the derivative of 4x^2+3 so that means you make a substitution for 4x^2+3. This works because in the substitution method you find the derivative of 4x^2+3 and this ends up "cancelling" with the 3x.

    Or the quickest method is to consider the derivative of (4x^2+3)^6 but this takes a lot of practice.
    Yeah I'm definitely still trying to get to grips with f'(x)[f(x)]^n = \frac{[f(x)]^{n+1}}{n+1} + c

    It may take me a while
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    (Original post by Philip-flop)
    Yeah I'm definitely still trying to get to grips with f'(x)[f(x)]^n = \frac{[f(x)]^{n+1}}{n+1} + c

    It may take me a while
    This is what inspection is, in essence.
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    (Original post by RDKGames)
    This is what inspection is, in essence.
    Yeah I'm slowly getting the hang of it! I remember when I first came across inspection of  \int \frac{f'(x)}{f(x)} types, I just didn't understand any of it, but I got there in the end
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    So I just attempted Question 5 of Edexcel C4 January 2013 paper and came across a few silly little things I have forgotten such as the fact that  \frac{d(a^x)}{dx} = a^x ln a

    But then I came across something that I can't recall ever seeing in my life!  \int a^x dx = \frac{a^x}{ln a}
    Why have I never come across this before?!!
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    (Original post by Philip-flop)
    So I just attempted Question 5 of Edexcel C4 January 2013 paper and came across a few silly little things I have forgotten such as the fact that  \frac{d(a^x)}{dx} = a^x ln a

    But then I came across something that I can't recall ever seeing in my life!  \int a^x dx = \frac{a^x}{ln a}
    Why have I never come across this before?!!
    They don't come up often, and I think they're in the formula booklet, so it doesn't matter whether you know them or not. Have a go at deriving it yourself so you how it's true and be able to derive it yourself in the exam.
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    (Original post by RDKGames)
    They don't come up often, and I think they're in the formula booklet, so it doesn't matter whether you know them or not. Have a go at deriving it yourself so you how it's true and be able to derive it yourself in the exam.
    They're not in the formula book for Edexcel (unless I'm being stupid). I know how to prove  \frac{d(a^x)}{dx} = a^x ln a but I have no idea how to prove  \int a^x dx = \frac{a^x}{ln a}

    Reality is starting to hit me that I'm not prepared for these exams over the next 2 months
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    (Original post by Philip-flop)
    They're not in the formula book for Edexcel (unless I'm being stupid). I know how to prove  \frac{d(a^x)}{dx} = a^x ln a but I have no idea how to prove  \int a^x dx = \frac{a^x}{ln a}

    Reality is starting to hit me that I'm not prepared for these exams over the next 2 months
    Just try different methods. Have a go, see what happens.
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    (Original post by Philip-flop)
    So I just attempted Question 5 of Edexcel C4 January 2013 paper and came across a few silly little things I have forgotten such as the fact that  \frac{d(a^x)}{dx} = a^x ln a

    But then I came across something that I can't recall ever seeing in my life!  \int a^x dx = \frac{a^x}{ln a}
    Why have I never come across this before?!!
    Yes when this integral came up it stumped a lot of students.

    If you know that the derivative of a^x is a^x \ln a then you should be able to reverse this to find the integral of a^x since \ln a is just a constant.

    Alternatively, you are given this result in the formula book:

    a^x = e^{x\ln a}

    You can use this to find both the derivative and integral of a^x. Let us know if you're not sure how.

    But for the derivative of a^x it's worth also knowing the proof that involves taking logs of both sides / implicit differentiation since there's a small chance that it would be required in the exam.
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    (Original post by notnek)
    Yes when this integral came up it stumped a lot of students.

    If you know that the derivative of a^x is a^x \ln a then you should be able to reverse this to find the integral of a^x since \ln a is just a constant.

    Alternatively, you are given this result in the formula book:

    a^x = e^{x\ln a}

    You can use this to find both the derivative and integral of a^x. Let us know if you're not sure how.

    But for the derivative of a^x it's worth also knowing the proof that involves taking logs of both sides / implicit differentiation since there's a small chance that it would be required in the exam.
    SHOULD be able to but I don't think I understand how I can prove  \frac{d(a^x)}{dx} = a^x ln a but I don't know how to prove  \int a^x dx = \frac{a^x}{ln a} FML why is everything so difficult for me to get my head around? :/

    Also I have no idea how a^x = e^{x\ln a} is linked to the above Seriously feel like giving up this close to the exams!
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    (Original post by Philip-flop)
    SHOULD be able to but I don't think I understand how I can prove  \frac{d(a^x)}{dx} = a^x ln a but I don't know how to prove  \int a^x dx = \frac{a^x}{ln a} FML why is everything so difficult for me to get my head around? :/

    Also I have no idea how a^x = e^{x\ln a} is linked to the above Seriously feel like giving up this close to the exams!
    Well... you know how to integrate e^{ax} where a is a constant, so you already know how to integrate e^{x\ln(a)}
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    (Original post by RDKGames)
    Well... you know how to integrate e^{ax} where a is a constant, so you already know how to integrate e^{x\ln(a)}
    So if...
     \int e^{2x} = \frac{1}{2} e^{2x} + c

    then...
     \int e^{xln(a)} = \frac{1}{ln(a)} e^{xln(a)} + c ??? Something tells me I did something wrong there!
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    (Original post by Philip-flop)
    So if...
     \int e^{2x} = \frac{1}{2} e^{2x}

    then...
     \int e^{xln(a)} = \frac{1}{ln(a)} e^{xln(a)} ??? Something tells me I did something wrong there!
    Why do you think its wrong?
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    (Original post by RDKGames)
    Why do you think its wrong?
    Cos it's not the same format as  \int a^x dx = \frac{a^x}{ln a}

    I actually have no idea what I'm talking about right now
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    (Original post by Philip-flop)
    Cos it's not the same format as  \int a^x dx = \frac{a^x}{ln a}

    I actually have no idea what I'm talking about right now
    Yes it is. Look at all the information, there is one equality that shows they're both the same.
 
 
 
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