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    "hi i like to post integrals way above the level of the C4 spec to let everyone know i have a massive dong"
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    (Mod hat on)

    This is Philip-flop's thread for help with C4 integration so please keep the thread relevant to him and any other C4 students visiting this thread. If you want to discuss harder / higher level integrals then there are threads in the index that are crying out for people to post in them!


    Any more off topic posts will be removed.
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    (Original post by notnek)
    I think I used to but I don't anymore sorry. You can always use Wolfram Alpha to check your answers or you could ask here.
    Philip-flop


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    \displaystyle \int x(3x+2)^2 .dx = \frac{1}{36}(3x+2)^4-\frac{2}{27}(3x+2)^3 +c

    \displaystyle \int 3x^{-1} .dx = 3\ln \vert x \vert +c

    \displaystyle \int \cos(2x-3).dx = \frac{1}{2}\sin(2x-3) +c

    \displaystyle \int e^{3x-7}.dx = \frac{1}{3}e^{3x-7} +c

    \displaystyle \int -2\csc(3x)\cot(3x) .dx = \frac{2}{3}\csc(3x) +c

    \displaystyle \int 2x(x^2-3)^4 .dx = \frac{1}{5}(x^2-3)^5 + c

    \displaystyle \int \frac{3x-6}{x^2} .dx = 3\ln \vert x \vert + \frac{6}{x} +c

    \displaystyle \int \frac{8x^3}{x^4-3} .dx =  2\ln  \vert x^4 -3 \vert + c

    \displaystyle \int \cos(3x-2)\sin^2(3x-2) .dx = \frac{1}{9}\sin^3(3x-2) +c

    \displaystyle \int 8xe^{x^2}.dx = 4e^{x^2} +c

    \displaystyle \int sec^3(x)\tan(x) .dx = \frac{1}{3}sec^3(x) +c

    \displaystyle \int \cos(x^2) .dx = eh... something too complicated for A-Level..

    \displaystyle \int  3\sin^2(2x) .dx = \frac{3}{2}x-\frac{3}{8}\sin(4x) +c

    \displaystyle \int \frac{3x+5}{3x^2+10x} .dx = \frac{1}{2}\ln \vert 3x^2+10x \vert +c

    \displaystyle \int 3\tan(5x) .dx = \frac{3}{5}\ln \vert \sec(5x) \vert +c

    \displaystyle \int 5+tan^2(x) .dx = 4x+tan(x) +c

    \displaystyle \int \frac{10}{9x^2-16} .dx = \frac{5}{12}\ln \left\vert \frac{3x-4}{3x+4} \right\vert +c

    \displaystyle \int x(x-2)^{10} .dx = \frac{1}{12}(x-2)^{12}+\frac{2}{11}(x-2)^{11}+c

    \displaystyle \int x\cos(x) .dx = x\sin(x)+\cos(x)+c

    \displaystyle \int \frac{3}{(x+1)(x-2)} .dx = \ln \left\vert \frac{x-2}{x+1} \right\vert +c

    \displaystyle \int \frac{2}{(3x-2)^2} .dx = \frac{2}{6-9x} +c

    \displaystyle \int \frac{\sqrt{x^2+4}}{x}.dx = eh... nothing that you have to worry about appearing in C4..

    \displaystyle \int x^2e^x .dx = e^x(x^2-2x+2) +c

    \displaystyle \int \frac{5}{\cos(x)}.dx = 5\ln \vert \sec(x) + \tan(x) \vert +c

    \displaystyle \int 5\ln(3x) .dx = 5x\ln (3x)-5x +c

    \displaystyle \int \sqrt{3x-8}.dx = \frac{2}{9}(3x-8)^{3/2}+c

    \displaystyle \int x^2\ln(x) .dx = \frac{x^3}{3}\ln(x)-\frac{x^3}{9} +c

    \displaystyle \int sin^3(x).dx = -\cos(x) +\frac{1}{3}cos^3(x) +c

    \displaystyle \int \sec(x)\tan(x)e^{\sec(x)}.dx = e^{\sec(x)} +c

    \displaystyle \int 3\sin(8x)\cos(8x).dx = \frac{3}{16}\sin^2(8x) +c_1 = -\frac{3}{16}\cos^2(8x)+c_2

    \displaystyle \int \frac{2x^2-3}{x^2-1}.dx = \frac{5}{2}\ln \vert x+1 \vert - \frac{1}{2}\ln \vert x-1 \vert +c


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    (Original post by RDKGames)
    Philip-flop


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    \displaystyle \int x(3x+2)^2 .dx = \frac{1}{36}(3x+2)^4-\frac{2}{27}(3x+2)^3 +c

    \displaystyle \int 3x^{-1} .dx = 3\ln \vert x \vert +c

    \displaystyle \int \cos(2x-3).dx = \frac{1}{2}sin(2x-3) +c

    \displaystyle \int e^{3x-7}.dx = \frac{1}{3}e^{3x-7} +c

    \displaystyle \int -2\csc(3x)\cot(3x) .dx = \frac{2}{3}\csc(3x) +c

    \displaystyle \int 2x(x^2-3)^4 .dx = \frac{1}{5}(x^2-3)^5 + c

    \displaystyle \int \frac{3x-6}{x^2} .dx = 3\ln \vert x \vert + \frac{6}{x} +c

    \displaystyle \int \frac{8x^3}{x^4-3} .dx =  2\ln  \vert x^4 -3 \vert + c

    \displaystyle \int \cos(3x-2)\sin^2(3x-2) .dx = \frac{1}{9}\sin^3(3x-2) +c

    \displaystyle \int 8xe^{x^2}.dx = 4e^{x^2} +c

    \displaystyle \int sec^3(x)\tan(x) .dx = \frac{1}{3}sec^3(x) +c

    \displaystyle \int \cos(x^2) .dx = eh... something too complicated for A-Level..

    \displaystyle \int  3\sin^2(2x) .dx = \frac{3}{2}x-\frac{3}{8}\sin(4x) +c

    \displaystyle \int \frac{3x+5}{3x^2+10x} .dx = \frac{1}{2}\ln \vert 3x^2+10x \vert +c

    \displaystyle \int 3\tan(5x) .dx = \frac{3}{5}\ln \vert \sec(5x) \vert +c

    \displaystyle \int 5+tan^2(x) .dx = 4x+tan(x) +c

    \displaystyle \int \frac{10}{9x^2-16} .dx = \frac{5}{12}\ln \left\vert \frac{3x-4}{3x+4} \right\vert +c

    \displaystyle \int x(x-2)^{10} .dx = \frac{1}{12}(x-2)^{12}+\frac{2}{11}(x-2)^{11}+c

    \displaystyle \int x\cos(x) .dx = x\sin(x)+\cos(x)+c

    \displaystyle \int \frac{3}{(x+1)(x-2)} .dx = \ln \left\vert \frac{x-2}{x+1} \right\vert +c

    \displaystyle \int \frac{2}{(3x-2)^2} .dx = \frac{2}{6-9x} +c

    \displaystyle \int \frac{\sqrt{x^2+4}}{x}.dx = eh... nothing that you have to worry about appearing in C4..

    \displaystyle \int x^2e^x .dx = e^x(x^2-2x+2) +c

    \displaystyle \int \frac{5}{\cos(x)}.dx = 5\ln \vert \sec(x) + \tan(x) \vert +c

    \displaystyle \int 5\ln(3x) .dx = 5x\ln (3x)-5x +c

    \displaystyle \int \sqrt{3x-8}.dx = \frac{2}{9}(3x-8)^{3/2}+c

    \displaystyle \int x^2\ln(x) .dx = \frac{x^3}{3}\ln(x)-\frac{x^3}{9} +c

    \displaystyle \int sin^3(x).dx = -\cos(x) +\frac{1}{3}cos^3(x) +c

    \displaystyle \int \sec(x)\tan(x)e^{\sec(x)}.dx = e^{\sec(x)} +c

    \displaystyle \int 3\sin(8x)\cos(8x).dx = \frac{3}{16}\sin^2(8x) +c

    \displaystyle \int \frac{2x^2-3}{x^2-1}.dx = \frac{5}{2}\ln \vert x+1 \vert - \frac{1}{2}\ln \vert x-1 \vert +c




    Thanks so much for that.

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    Yes one of them can't be expressed in terms of elementary functions - when I use this list I check that the student doesn't try and use reverse chain rule for this. The other one you singled out isn't C4 so I'm not sure what it's doing there


    Philip-flop I forgot to say that this is a list of integrals where the method isn't given to you. Since for substitution questions the substitutions will always be given to you in C4, there are no substitution integrals in this list (well none that require substitution).
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    (Original post by notnek)
    Thanks so much for that.

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    Yes one of them can't be expressed in terms of elementary functions - when I use this list I check that the student doesn't try and use reverse chain rule for this. The other one you singled out isn't C4 so I'm not sure what it's doing there



    Philip-flop I forgot to say that this is a list of integrals where the method isn't given to you. Since for substitution questions the substitutions will always be given to you in C4, there are no substitution integrals in this list (well none that require substitution).
    Maybe that \cos(x^2) integral was supposed to be \cos(\sqrt{x})?? In which case solving it would require a nice mix of two different methods.
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    (Original post by notnek)
    Thanks so much for that.

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    Yes one of them can't be expressed in terms of elementary functions - when I use this list I check that the student doesn't try and use reverse chain rule for this. The other one you singled out isn't C4 so I'm not sure what it's doing there



    Philip-flop I forgot to say that this is a list of integrals where the method isn't given to you. Since for substitution questions the substitutions will always be given to you in C4, there are no substitution integrals in this list (well none that require substitution).
    That awkward moment when your answer to the first integral was worked out by using Substitution I'm having a lot of trouble recognising what type of integration to use
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    (Original post by Philip-flop)
    That awkward moment when your answer to the first integral was worked out by using Substitution I'm having a lot of trouble recognising what type of integration to use
    The first one can be done simply by expanding the brackets. Always look for the simplest method first
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    (Original post by RDKGames)
    Maybe that \cos(x^2) integral was supposed to be \cos(\sqrt{x})?? In which case solving it would require a nice mix of two different methods.
    No it was meant to be \cos(x^2) to make sure that students don't try to use some kind of reverse chain rule.
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    (Original post by notnek)
    No it was meant to be \cos(x^2) to make sure that students don't try to use some kind of reverse chain rule.
    Oh fair enough. Also the same integral pops up twice in there, \int \frac{3}{(x+1)(x-2)}.dx
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    (Original post by notnek)
    The first one can be done simply by expanding the brackets. Always look for the simplest method first
    Oh I see! I never would have known to do that!
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    (Original post by Philip-flop)
    Oh I see! I never would have known to do that!
    There are a few in there that are C1 integrals but they're mixed in with C4 integrals, which is why students often forget the easy methods.
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    (Original post by notnek)
    There are a few in there that are C1 integrals but they're mixed in with C4 integrals, which is why students often forget the easy methods.
    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

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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

    This one is much easier than you think. No int by parts needed.
    How does the 2x relate to the thing inside the brackets?
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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 can certainly use substitution, but you can use inspection for that one.
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    Oh yeah it's in the format  \int f'(x) [f(x)]^n dx

    But these ones I'm still not use to doing

    EDIT: I managed to get it but only by looking at the general formula for these types. I would have had to resort to Substitution if I was given this in a real exam!
    (Original post by carpetguy)
    This one is much easier than you think. No int by parts needed.
    How does the 2x relate to the thing inside the brackets?
    (Original post by RDKGames)
    You can certainly use substitution, but you can use inspection for that one.
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    (Original post by Philip-flop)
    EDIT: I managed to get it but only by looking at the general formula for these types. I would have had to resort to Substitution if I was given this in a real exam!
    That's fine, as long as you solve it and don't take up too much time...
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    (Original post by Philip-flop)
    Oh yeah it's in the format  \int f'(x) [f(x)]^n dx

    But these ones I'm still not use to doing
    When looking at an integral that one there should be the first or second thing you look for. Once you get more familiar with all the methods the easiest way to do integration is looking and seeing if you can do the easiest stuff first. Start with "can I just expand it or something", then this thing:


    and so on. You can usually attempt a substitution for most if all else fails, but if you're like me and are a bit :innocent::innocent::innocent::innocent: at the whole C4 shabang it's generally an awful idea unless the question specifically gives you a substitution to use.
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    (Original post by RDKGames)
    That's fine, as long as you solve it and don't take up too much time...
    Yeah, Integration by Substitution will hopefully be a last resort when doing the exam! I've just got to try and think strategically I guess.

    (Original post by Retired_Messiah)
    When looking at an integral that one there should be the first or second thing you look for. Once you get more familiar with all the methods the easiest way to do integration is looking and seeing if you can do the easiest stuff first. Start with "can I just expand it or something", then this thing:


    and so on. You can usually attempt a substitution for most if all else fails, but if you're like me and are a bit :innocent::innocent::innocent::innocent: at the whole C4 shabang it's generally an awful idea unless the question specifically gives you a substitution to use.
    Useful advice. I will definitely try to tackle integrals in a more systematic way by thinking of the easiest routes first and leave Substitution as a last resort.
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    Can someone explain how I'm meant to start this?...
     \int cos(3x-2)sin^2(3x-2) dx
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    (Original post by Philip-flop)
    Can someone explain how I'm meant to start this?...
     \int cos(3x-2)sin^2(3x-2) dx
    Substitute u=\sin(3x-2)

    Or recognise that it's almost in the form f'(x)[f(x)]^n
 
 
 
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