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Integration by substitution

  1. Nov 15, 2011 #1
    1. The problem statement, all variables and given/known data
    I have the integral
    [tex]\int sin(2.13\sqrt{x}+2.4)\,dx[/tex]

    I'm supposed to use the substitution [tex]y=2.13\sqrt{x}+2.4[/tex], aka. [tex]\sqrt{x}=\frac {y-2.4}{2.13}[/tex] to gain the following description of the integral:

    [tex]\int sin(2.13\sqrt{x}+2.4)\,dx = E cos(y) + F\int y sin(y)\,dy[/tex]

    I have obviously not tried every method possible, but I'm running out of ideas on how to proceed on this one. I tried using the quotient rule without any good looking results, probably due to my non-existing knowledge about these mathematical methods. Any help would be appreciated!
     
  2. jcsd
  3. Nov 15, 2011 #2

    I like Serena

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    Welcome to PF, LizzieL! :smile:

    What you need to do is "Integration by substitution".

    First you need an expression for x (the square of what you have).
    Then you need to differentiate it to find dx.

    If you have that you need to replace the parts in your original integral that contain x, by the parts that contain y.
    In particular dx needs to be replaced by the derivative you should have for x, followed by "dy".
     
  4. Nov 15, 2011 #3
    Thank you!

    Ok, so I have

    [tex]x=\big( \frac {y-2.4}{2.13}\big)^2[/tex]

    And,
    [tex]dx= 2 \big( \frac {y-2.4}{2.13}\big) dy[/tex]

    Is this what you mean? I don't think I understood the last part you mentioned though.
     
  5. Nov 15, 2011 #4

    I like Serena

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    Yes, this is what I meant! :smile:

    However, you did not properly apply the chain rule yet.
    Do you know what the chain rule is? And more importantly, how to apply it?



    As for the last part I mentioned, it was that in your integral you should replace:

    [itex](2.13\sqrt{x}+2.4)[/itex] by [itex]y[/itex],

    and you should replace:

    [itex]dx[/itex] by what you just found (after you correct it for the application of the chain rule).
     
  6. Nov 15, 2011 #5
    Normally I don't think I have a problem understanding the chain rule, but it's all backwards when in integration context, and it confuses me.
    My textbook is not covering this topic properly (in my opinion). So to answer your question, I do not know how to apply the chain rule here... :shy:
     
  7. Nov 15, 2011 #6

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    Well, in this case it is not backward...
    You need to multiply your result for dx with the derivative of [itex](\frac {y-2.4}{2.13})[/itex].

    Aaaand... I'm off to bed now! :zzz:
     
  8. Nov 16, 2011 #7
    I'm stuck! :cry:
    I'm desperately trying to figure out the derivative of
    [tex]y=2.13\sqrt{x}+2.4[/tex]

    but I can't seem to get it right.
    I'm starting off like this:

    [tex]y'=(2.13\sqrt{x}+2.4)'= 2.13x^{1/2} ???[/tex]

    How do I calculate this?
     
  9. Nov 16, 2011 #8

    Mark44

    Staff: Mentor

    Write your first equation using exponents rather than radicals.
    y = 2.13 x1/2 + 2.4

    Now, use the power rule to find y'.
     
  10. Nov 16, 2011 #9
    Ok, thanks!
    So, is this right or close to right?:

    [tex]y^,= 2.13x^{1/2}= 2.13(\frac{1}{2})x^{-1/2}=\frac{2.13}{2\sqrt{x}}[/tex]

    And proceeding,

    [tex]dy= \frac{2.13}{2\sqrt{x}} dx[/tex] ?

    How do I move from here?
     
  11. Nov 16, 2011 #10

    Mark44

    Staff: Mentor

    It's partly right, but you are munging a bunch of stuff together that should be there.

    y = 2.13x1.2 + 2.4
    so y' = dy/dx = (2.13/2)x-1/2

    Don't confuse y with y' - they are different things.

    Now solve the equation above for dx, which you'll have to replace in your original integral. The original integral looks like this: [itex]\int f(x) dx[/itex]. Use your substitution to replace x and dx with y and dy to get an integral that looks like this: [itex]\int g(y) dy[/itex]. The goal of substitution is to get a different integral that is easier to calculate.
     
  12. Nov 16, 2011 #11
    Thank you for your help!

    I think I'm getting there:

    [tex]\frac{dy}{dx}= \frac {2.13}{2\sqrt{x}} \Rightarrow
    dx = \frac {2\sqrt{x}}{2.13}dy

    \Rightarrow
    \int sin(y) \cdot \frac {2}{2.13}\Big(\frac {y-2.4}{2.13}\Big)dy[/tex]

    [tex]= \frac {-2.4}{2.13^2} cos(y) + \frac {2}{2.13^2}\int y sin(y) dy[/tex]

    But I can't seem to get that last part right...the first coeffisient is correct, but not the second one. What did I do wrong?
     
  13. Nov 16, 2011 #12

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    Sorry, nevermind.
    Your problem is not asking you to do integration by parts.
    (Deleted previous post.)
     
  14. Nov 16, 2011 #13

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    Your second coefficient is correct.
    Why do you think it isn't?

    However, your first coefficient is not correct, due to the minus sign.
     
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