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Math Help - Integral 2

  1. #1
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    Integral 2

    Calculate:


    \int \frac{2dx}{\sqrt{2+x-x^2}}
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Apprentice123 View Post
    Calculate:


    \int \frac{2dx}{\sqrt{2+x-x^2}}
    2+x-x^2=2-(x^2-x)=2-\left(\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\right)=\frac{9}{4}-\left(x-\frac{1}{2}\right)^2

    Let x-\frac{1}{2}=\frac{3}{2}\sin(\theta)
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by mr fantastic View Post
    2 + x - x^2 < 0 for all real values of x so the integrand makes no sense.
    2+x-x^2>0

    Solving this we get

    -1<x<2
    .......
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    2+x-x^2=2-(x^2-x)=2-\left(\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\right)=\frac{9}{4}-\left(x-\frac{1}{2}\right)^2

    Let x-\frac{1}{2}=\sin(\theta)
    you mean x - \frac 12 = \frac 32 \sin \theta
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Jhevon View Post
    you mean x - \frac 12 = \frac 32 \sin \theta
    Yeah, thank you...forgot to type that part
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  6. #6
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    Quote Originally Posted by Mathstud28 View Post
    2+x-x^2>0

    Solving this we get

    -1<x<2
    .......
    *Ahem* Beat you to it (which is why the reply was deleted).
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  7. #7
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    I use that formula?

    \int \frac{du}{\sqrt{u^2+a^2}}=ln(u+\sqrt{u^2+a^2})


    How to solve?
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  8. #8
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    Quote Originally Posted by Apprentice123 View Post
    I use that formula?

    \int \frac{du}{\sqrt{u^2+a^2}}=ln(u+\sqrt{u^2+a^2})


    How to solve?
    \int \frac{2}{\sqrt{2+x-x^2}} \ \mathrm{d}x = 2 \int \frac{1}{\sqrt{2+x-x^2}} \ \mathrm{d}x =  2 \int \frac{1}{\sqrt{\frac{9}{4} - \left(x-\frac12\right)^2}} \ \mathrm{d}x  = 2 \int \frac{1}{\sqrt{\left(\frac{3}{2}\right)^2 - \left(x-\frac12\right)^2}} \ \mathrm{d}x

    2 \int \frac{1}{\sqrt{\left(\frac{3}{2}\right)^2 - \left(x-\frac12\right)^2}} \ \mathrm{d}x is in the form \int \frac{1}{\sqrt{a^2 - u^2}} \ \mathrm{d}u so can you continue?

    EDIT: Use this formula - \int \frac{1}{\sqrt{a^2-u^2}} \ \mathrm{d}u = \mathrm{arcsin}\left(\frac{u}{a}\right)
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  9. #9
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    This way?


    \frac{1}{2}arcsin(\frac{(x-\frac{1}{2})}{\frac{3}{2}})
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  10. #10
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    Quote Originally Posted by Apprentice123 View Post
    This way?


    \frac{1}{2}arcsin(\frac{(x-\frac{1}{2})}{\frac{3}{2}})
    You forgot the 2 which was outside the integral (the numerator).

    Also, the general rule is \int \frac{1}{\sqrt{a^2-u^2}} \ \mathrm{d}u = \mathrm{arcsin}\left(\frac{u}{a}\right) hence you do not need \frac12 at the start.

    Therefore the answer is:

    2 \int \frac{1}{\sqrt{\left(\frac{3}{2}\right)^2 - \left(x-\frac12\right)^2}} \ \mathrm{d}x = 2\left[\mathrm{arcsin}\left(\frac{\left(x-\frac12\right)}{\left(\frac32\right)}<br />
\right) \right]=  2\left[\mathrm{arcsin}\left(\frac{2\left (x-\frac12\right)}{3}<br />
\right)\right]= 2\left[\mathrm{arcsin}\left(\frac{\left (2x-1\right)}{3}<br />
 \right)\right] = 2\mathrm{arcsin}\left(\frac{2x-1}{3}<br />
 \right)
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  11. #11
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    thank you
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