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    \displaystyle\int^4_0 \frac{1}{\sqrt(9x^2+4)} \ dx .

    The root is underneath the entire denominator. I can do the problem I used a u-substitution u =3x. The only issue I have is that when I go to simplify it in logarithmic form, I've changed the limits from 4 and 0 to 12 and 0 respectively, but in the answer scheme the still have it as 4 and 0. Confused as to why the limits don't change.
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    (Original post by Music99)
    \displaystyle\int^4_0 \frac{1}{\sqrt(9x^2+4)} \ dx .

    The root is underneath the entire denominator. I can do the problem I used a u-substitution u =3x. The only issue I have is that when I go to simplify it in logarithmic form, I've changed the limits from 4 and 0 to 12 and 0 respectively, but in the answer scheme the still have it as 4 and 0. Confused as to why the limits don't change.
    Does the mark scheme convert the final answer back to a function of x before subbing in the limits? You use the limits that apply to the appropriate variable, so you

    EITHER put u = something, change limits, get out a function of u, put in u-limits

    OR put u = something, write limits as x = original limits, get out a function of u, convert back to a function of x, put in original x limits
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    (Original post by Music99)
    \displaystyle\int^4_0 \frac{1}{\sqrt(9x^2+4)} \ dx .

    The root is underneath the entire denominator. I can do the problem I used a u-substitution u =3x. The only issue I have is that when I go to simplify it in logarithmic form, I've changed the limits from 4 and 0 to 12 and 0 respectively, but in the answer scheme the still have it as 4 and 0. Confused as to why the limits don't change.
    What do you mean by logarithmic form, exactly? There's nothing of the sort involved in this question.

    But yes, for the substitution u=3x, you must change the limits from 4 and 0 to 12 and 0. If the answer scheme says otherwise, it's wrong.
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    (Original post by justinawe)
    What do you mean by logarithmic form, exactly? There's nothing of the sort involved in this question.

    But yes, for the substitution u=3x, you must change the limits from 4 and 0 to 12 and 0. If the answer scheme says otherwise, it's wrong.
    so expressing the cosh function in it's logarithmic form.
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    (Original post by davros)
    Does the mark scheme convert the final answer back to a function of x before subbing in the limits? You use the limits that apply to the appropriate variable, so you

    EITHER put u = something, change limits, get out a function of u, put in u-limits

    OR put u = something, write limits as x = original limits, get out a function of u, convert back to a function of x, put in original x limits
    I'm not to sure what they do. It's a question from the textbook so only has the answer, I imagine from what you have said they do the first choice. Okay so then I can just change my limits to 12 and 0, and then just sub those into the logarithmic expression and that's my answer?
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    (Original post by Music99)
    so expressing the cosh function in it's logarithmic form.
    Oh right.

    I'm pretty sure \displaystyle \int \frac{1}{\sqrt{x^2 + a^2}} \mathrm { \ } dx = \mathrm{arsinh} \left( \frac{x}{a} \right) , though? Not arcosh.
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    (Original post by Music99)
    I'm not to sure what they do. It's a question from the textbook so only has the answer, I imagine from what you have said they do the first choice. Okay so then I can just change my limits to 12 and 0, and then just sub those into the logarithmic expression and that's my answer?
    If you make the substitution u = 3x then yes, your u-limits are 0 and 12, so if you integrate to get a new function F(u) you work out F(12) - F(0). It doesn't matter whether you leave it as a hyperbolic form or a log form. Does your final answer agree with the book's answer?
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    (Original post by Music99)
    \displaystyle\int^4_0 \frac{1}{\sqrt(9x^2+4)} \ dx .

    The root is underneath the entire denominator. I can do the problem I used a u-substitution u =3x. The only issue I have is that when I go to simplify it in logarithmic form, I've changed the limits from 4 and 0 to 12 and 0 respectively, but in the answer scheme the still have it as 4 and 0. Confused as to why the limits don't change.
    take the factor of 2 outside the square root so now you have sqrt(9x^2/4 +1) then use the substitution sinh(u)=3x/2 no idea what your trying to do with 'logarithmic form'
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    (Original post by justinawe)
    Oh right.

    I'm pretty sure \displaystyle \int \frac{1}{\sqrt{x^2 + a^2}} \mathrm { \ } dx = \mathrm{arsinh} \left( \frac{x}{a} \right) , though? Not arcosh.
    DOH!!
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    (Original post by natninja)
    take the factor of 2 outside the square root so now you have sqrt(9x^2/4 +1) then use the substitution sinh(u)=3x/2 no idea what your trying to do with 'logarithmic form'
    The question asks to simplify it into log form.
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    (Original post by Music99)
    but in the answer scheme the still have it as 4 and 0. Confused as to why the limits don't change.[/FONT][/COLOR]
    so only has the answer
    This is rather a confusing thread, with information only coming to light as people respond to you. You've posted on here a fair bit, so have a bit more thought as to what's relevant when starting your thread; pretty please.
 
 
 
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