Integration by substitution

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  1. JeremyB's Avatar
    1 28-05-2012  19:41
    • Junior Member
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    Integration by substitution
    Turns out I'm finding integration by substitution the most challenging part of the whole A-level. I'm not sure if its meant to be; I'm probably missing something.

    But I really cannot solve this at all:

    \displaystyle \int \dfrac{(x^2 + 4)^{1/2}}{x}\ dx using u^2= x^2 + 4.

    The answer is supposed to be:

    (x^2 + 4)^{1/2} + \ln \dfrac{(x^2 + 4)^{1/2}-2}{(x^2 + 4)^{1/2} + 2}

    So here's what I do:

    x = (u^2 + 4)^{1/2}, \\\ 2u\dfrac{du}{dx} = 2x, \\\ \dfrac{u}{(u^2 + 4)^{1/2}} \dfrac{du}{dx} = 1

    Then the integral becomes:

    \displaystyle \int \dfrac{u^2}{u^2 + 4}\ du.

    which to me seems unsolvable. So I must have done something wrong but I can't imagine how I could do it differently. I've tried about ten times, so could someone please help me. I would really appreciate it.
    Last edited by JeremyB; 28-05-2012 at 19:42.
  2. ecraven's Avatar
    2 28-05-2012  19:59
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    Re: Integration by substitution
    Wouldn't the first part of your method be x = (u^2 - 4)^{1/2} instead?
  3. BabyMaths's Avatar
    3 28-05-2012  20:00
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    Re: Integration by substitution
    (Original post by JeremyB)

    Then the integral becomes:

    \displaystyle \int \dfrac{u^2}{u^2 + 4}\ du.
    I didn't check you whole post but

    \displaystyle \int \dfrac{u^2}{u^2 + 4}\ du=\int \dfrac{u^2+4-4}{u^2 + 4}\ du=\int 1 - \frac{4}{u^2 + 4} du.

    The second term is a standard inegral.
  4. JeremyB's Avatar
    4 28-05-2012  20:08
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    Re: Integration by substitution
    (Original post by ecraven)
    Wouldn't the first part of your method be x = (u^2 - 4)^{1/2} instead?
    Oops yes that's what I got but I copied it wrong on the computer.
  5. raheem94's Avatar
    5 28-05-2012  20:11
    • TSR Demigod
    • Posts: 5,512
    Re: Integration by substitution
    (Original post by JeremyB)
    Turns out I'm finding integration by substitution the most challenging part of the whole A-level. I'm not sure if its meant to be; I'm probably missing something.

    But I really cannot solve this at all:

    \displaystyle \int \dfrac{(x^2 + 4)^{1/2}}{x}\ dx using u^2= x^2 + 4.

    The answer is supposed to be:

    (x^2 + 4)^{1/2} + \ln \dfrac{(x^2 + 4)^{1/2}-2}{(x^2 + 4)^{1/2} + 2}

    So here's what I do:

    x = (u^2 + 4)^{1/2}, \\\ 2u\dfrac{du}{dx} = 2x, \\\ \dfrac{u}{(u^2 + 4)^{1/2}} \dfrac{du}{dx} = 1

    Then the integral becomes:

    \displaystyle \int \dfrac{u^2}{u^2 + 4}\ du.

    which to me seems unsolvable. So I must have done something wrong but I can't imagine how I could do it differently. I've tried about ten times, so could someone please help me. I would really appreciate it.
    \displaystyle dx = \frac{udu}{x} \ \ \ \ \ x^2 = u^2-4

    \displaystyle \int \frac{(x^2+4)^{\frac12}}{x} \ dx = \int \frac{u}{x} \times \frac{udu}{x} \ du = \int \frac{u^2}{x^2} \ du = \int \frac{u^2}{u^2-4} \ du \\ = \int \frac{u^2-4+4}{u^2-4} \ du = \int \left( \frac{u^2-4}{u^2-4} + \frac4{u^2-4} \right) \ du = \int \left( 1 + \frac4{u^2-4} \right) \ du \\ = \int \left( 1 + \frac4{(u-2)(u+2)} \right) \ du

    Use partial fraction on \displaystyle \frac4{(u-2)(u+2)}
  6. JeremyB's Avatar
    6 28-05-2012  20:12
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    Re: Integration by substitution
    (Original post by BabyMaths)
    I didn't check you whole post but

    \displaystyle \int \dfrac{u^2}{u^2 + 4}\ du=\int \dfrac{u^2+4-4}{u^2 + 4}\ du=\int 1 - \frac{4}{u^2 + 4} du.

    The second term is a standard inegral.
    Thanks a lot for that. Simple enough once you see it.
  7. raheem94's Avatar
    7 28-05-2012  20:13
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    Re: Integration by substitution
    (Original post by BabyMaths)
    I didn't check you whole post but

    \displaystyle \int \dfrac{u^2}{u^2 + 4}\ du=\int \dfrac{u^2+4-4}{u^2 + 4}\ du=\int 1 - \frac{4}{u^2 + 4} du.

    The second term is a standard inegral.
    As the OP is struggling on integration by substitution(C4 topic), so i am quite sure that he doesn't knows the standard integral you are talking about. He is probably working at C4 not FP3, this integral is in FP3.

    Actually he has probably made a sign error.
  8. JeremyB's Avatar
    8 28-05-2012  20:14
    • Junior Member
    • Posts: 28
    Re: Integration by substitution
    (Original post by raheem94)
    \displaystyle dx = \frac{udu}{x} \ \ \ \ \ x^2 = u^2-4

    \displaystyle \int \frac{(x^2+4)^{\frac12}}{x} \ dx = \int \frac{u}{x} \times \frac{udu}{x} \ du = \int \frac{u^2}{x^2} \ du = \int \frac{u^2}{u^2-4} \ du \\ = \int \frac{u^2-4+4}{u^2-4} \ du = \int \left( \frac{u^2-4}{u^2-4} + \frac4{u^2-4} \right) \ du = \int \left( 1 + \frac4{u^2-4} \right) \ du \\ = \int \left( 1 + \frac4{(u-2)(u+2)} \right) \ du

    Use partial fraction on \displaystyle \frac4{(u-2)(u+2)}
    Thanks mate. I can see it now.
  9. raheem94's Avatar
    9 28-05-2012  20:15
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    Re: Integration by substitution
    (Original post by JeremyB)
    Thanks a lot for that. Simple enough once you see it.
    Do you really know how to integrate \displaystyle \frac{4}{u^2 + 4} \ du \ ???

    I thought you are working at C4 :confused:
  10. JeremyB's Avatar
    10 28-05-2012  20:22
    • Junior Member
    • Posts: 28
    Re: Integration by substitution
    (Original post by raheem94)
    Do you really know how to integrate \displaystyle \frac{4}{u^2 + 4} \ du \ ???

    I thought you are working at C4 :confused:
    Yeah I wrote that one wrong on the computer when I was copying it from my notes.

    What I've been getting is u^2 / (u^2 - 4), which as you said I would do with partial fractions once simplified. My problem was the algebra involved in simplifying it first.

    Have you got any tips for solving these integration by substituion problems? I don't know why I'm finding it so tricky.
    Last edited by JeremyB; 28-05-2012 at 20:25.
  11. raheem94's Avatar
    11 28-05-2012  20:29
    • TSR Demigod
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    Re: Integration by substitution
    (Original post by JeremyB)
    Yeah I wrote that one wrong on the computer when I was copying it from my notes.

    What I've been getting is u^2 / (u^2 - 4), which as you said I would do with partial fractions once simplified. My problem was the algebra involved in simplifying it first.

    Have you got any tips for solving these integration by substituion problems? I don't know why I'm finding it so tricky.
    I don't have many tips for it, just practice more, with practice you will become better at recognizing the next steps.

    At first it looks difficult, but soon, with more practice, it becomes very easy.
  12. BabyMaths's Avatar
    12 28-05-2012  23:39
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    Re: Integration by substitution
    (Original post by raheem94)
    As the OP is struggling on integration by substitution(C4 topic), so i am quite sure that he doesn't knows the standard integral you are talking about. He is probably working at C4 not FP3, this integral is in FP3.

    Actually he has probably made a sign error.
    I believe you are correct.

    Sorry.
  13. raheem94's Avatar
    13 28-05-2012  23:50
    • TSR Demigod
    • Posts: 5,512
    Re: Integration by substitution
    (Original post by BabyMaths)
    I believe you are correct.

    Sorry.
    No problem.
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