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Thread: Integral 2

  1. #1
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    Integral 2

    Calculate:


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


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

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

    Solving this we get

    $\displaystyle -1<x<2$
    .......
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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
    $\displaystyle 2+x-x^2=2-(x^2-x)=2-\left(\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\right)=\frac{9}{4}-\left(x-\frac{1}{2}\right)^2$

    Let $\displaystyle x-\frac{1}{2}=\sin(\theta)$
    you mean $\displaystyle x - \frac 12 = \frac 32 \sin \theta$
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Jhevon View Post
    you mean $\displaystyle x - \frac 12 = \frac 32 \sin \theta$
    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
    $\displaystyle 2+x-x^2>0$

    Solving this we get

    $\displaystyle -1<x<2$
    .......
    *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?

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


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

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


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

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

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


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


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

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

    Therefore the answer is:

    $\displaystyle 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)}
    \right) \right]=$$\displaystyle 2\left[\mathrm{arcsin}\left(\frac{2\left (x-\frac12\right)}{3}
    \right)\right]= 2\left[\mathrm{arcsin}\left(\frac{\left (2x-1\right)}{3}
    \right)\right] = 2\mathrm{arcsin}\left(\frac{2x-1}{3}
    \right)$
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  11. #11
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    thank you
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