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     <font color="#444444"><span style="font-family: verdana">\displaystyle\int^\frac{\sqrt3}{  4}_0 \frac{2xsin^{-1}2x}{\sqrt(1-4x^2)}\ dx .

    The square root in the denominator is over everything. So In the previous part of the question I worked out that the derivative of  y= x-\sqrt(1-x^2)sin^{-1}x is  \frac{xsin^{-1}x}{\sqrt(1-x^2)} .

    so I called the  y= x-\sqrt(1-x^2)sin^{-1}x f(x), so then what I want to integrate is essentially f(2x), so then I just replaced all the x values in y=... by 2x, but then when I substitute in the limits I get the wrong answer. Any hints?


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    (Original post by Music99)
    Any hints?
    Do you get twice what the answer should be by any chance?
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    (Original post by Music99)
     <font color="#444444"><span style="font-family: verdana">\displaystyle\int^\frac{\sqrt3}{  4}_0 \frac{2xsin^{-1}2x}{\sqrt(1-4x^2)}\ dx .

    The square root in the denominator is over everything. So In the previous part of the question I worked out that the derivative of  y= x-\sqrt(1-x^2)sin^{-1}x is  \frac{xsin^{-1}x}{\sqrt(1-x^2)} .

    so I called the  y= x-\sqrt(1-x^2)sin^{-1}x f(x), so then what I want to integrate is essentially f(2x), so then I just replaced all the x values in y=... by 2x, but then when I substitute in the limits I get the wrong answer. Any hints?



    Try a substitution that gets your integral in the form of the derivative you found.
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    (Original post by ghostwalker)
    Do you get twice what the answer should be by any chance?
    I end up with  \frac{\sqrt3}{2}-\frac{\sqrt10}{4}(\frac{\pi}{3}) . the root 10 bit seems really out of place, but I double checked it and it gives the correct value for what I have as the integral.
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    (Original post by Music99)
    I end up with  \frac{\sqrt3}{2}-\frac{\sqrt10}{4}(\frac{\pi}{3}) . the root 10 bit seems really out of place, but I double checked it and it gives the correct value for what I have as the integral.
    I got \dfrac{\sqrt{3}}{4}-\dfrac{\pi}{12}


    What's the correct answer?
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    (Original post by ghostwalker)
    I got \dfrac{\sqrt{3}}{4}-\dfrac{\pi}{12}


    What's the correct answer?
    That is.
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    (Original post by ghostwalker)
    I got \dfrac{\sqrt{3}}{4}-\dfrac{\pi}{12}


    What's the correct answer?
    That is. Maybe I just subbed it in wrong or something, I'm confused now :P.
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    (Original post by BabyMaths)
    That is.
    Thanks.

    (Original post by Music99)
    That is. Maybe I just subbed it in wrong or something, I'm confused now :P.
    Post working?
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    (Original post by ghostwalker)
    Thanks.



    Post working?
    im on my phone so can't latex, but what I did was in the equation with y=... Everywhere there is an x I replaced it with a 2x then subbed in the limits.
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    (Original post by Music99)
    im on my phone so can't latex, but what I did was in the equation with y=... Everywhere there is an x I replaced it with a 2x then subbed in the limits.
    Safest way would be to use a subsitution, u=2x, so du/dx =2, and convert the limits. Then you'll have the exact form for the integral.
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    (Original post by ghostwalker)
    Safest way would be to use a subsitution, u=2x, so du/dx =2, and convert the limits. Then you'll have the exact form for the integral.
    Ah okay, also on questions like this one how do you know what substitution to use? I mean I know it comes from experience but is there a general rule of thumb.
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    (Original post by Music99)
    Ah okay, also on questions like this one how do you know what substitution to use? I mean I know it comes from experience but is there a general rule of thumb.
    Just compare corresponding parts of the formulae, and it should be obvious. You were aware of it when you initially thought of replacing x with 2x.

    It may be an interesting exercise to see how your original methodology didn't give the correct answer,
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    (Original post by Music99)
    I end up with  \frac{\sqrt3}{2}-\frac{\sqrt10}{4}(\frac{\pi}{3}) . the root 10 bit seems really out of place, but I double checked it and it gives the correct value for what I have as the integral.
    I'm struggling to see how you can get a \sqrt{10} factor in there!

    If you follow the advice given and set u = 2x then your upper integration limit becomes \sqrt{3}/2 so when you work out \sqrt{1 - u^2} this just gives you a factor 1/2. If you'd accidentally left in \sqrt{3}/4, you'd be taking the square root of 13/16, so I'm struggling to manufacture a 10 from somewhere
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    (Original post by davros)
    I'm struggling to see how you can get a \sqrt{10} factor in there!

    If you follow the advice given and set u = 2x then your upper integration limit becomes \sqrt{3}/2 so when you work out \sqrt{1 - u^2} this just gives you a factor 1/2. If you'd accidentally left in \sqrt{3}/4, you'd be taking the square root of 13/16, so I'm struggling to manufacture a 10 from somewhere
    I think I must have typed it in wrong on the calculator or something.
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    (Original post by Music99)
    I think I must have typed it in wrong on the calculator or something.
    OK, are you happy about following the suggestion made by Ghostwalker and how to get the correct answer now?
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    (Original post by davros)
    OK, are you happy about following the suggestion made by Ghostwalker and how to get the correct answer now?
    Yes .
 
 
 
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