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Math Help - root integral question

  1. #1
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    root integral question

    how to solve \int\sqrt{a-x^{2}}dx
    i tried by parts but its not not working
    any ideas
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  2. #2
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    Re: root integral question

    Quote Originally Posted by transgalactic View Post
    how to solve \int\sqrt{a-x^{2}}dx
    i tried by parts but its not not working any ideas
    Look at this.
    Last edited by Plato; June 25th 2011 at 06:18 AM.
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  3. #3
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    Re: root integral question

    what method they used?
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  4. #4
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    Re: root integral question

    Quote Originally Posted by transgalactic View Post
    what method they used?
    Click the button "show steps" at the right of the box.
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  5. #5
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    Re: root integral question

    See this. If you can calculate or know

    \int\frac{1}{\sqrt{a-x^2}}\;{dx}

    then doing the integral is very easy.
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  6. #6
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    Re: root integral question

    Quote Originally Posted by TheCoffeeMachine View Post
    See this. If you can calculate or know

    \int\frac{1}{\sqrt{a-x^2}}\;{dx}

    then doing the integral is very easy.
    In any 'manual' You can find the 'derivative' ...

    \frac{d}{dx} \sin^{-1} x = \frac{1}{\sqrt{1-x^{2}}} (1)

    ... so that is...

    \frac{d}{dx} \sin^{-1} \frac{x}{a} = \frac{1}{\sqrt{a^{2}-x^{2}}} (2)

    On WolframAlpha [I wonder why...] a different and more complicated formula is reported...

    Wolfram Mathematica Online Integrator

    Anyway following TCM we obtain...

    \int \sqrt {a^{2}-x^{2}}\ dx = \frac{x}{2}\ \sqrt{a^{2}-x^{2}} + \frac{a^{2}}{2}\ \sin^{-1} \frac {x}{a} + c (3)

    Kind regards

    \chi \sigma
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  7. #7
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    Re: root integral question

    Hello, transgalactic!

    \text{Integrate: }\:I \;=\;\int\sqrt{a-x^2}\:dx

    Use "Trig Substitution".
    Evidently, you aren't familiar with it . . .

    Let: x = a\sin\theta \quad\Rightarrow\quad dx \,=\,a\cos\theta\,d\theta

    Note that: \sqrt{a^2-x^2} \:=\:\sqrt{a^2-a^2\sin^2\!\theta} \:=\:\sqrt{a^2(1-\sin^2\!\theta)}
    . . . . . . . . =\;\sqrt{a^2\cos^2\!\theta} \;=\;a\cos\theta

    Substitute: . I \;=\;\int a\cos\theta(a\cos\theta\,d\theta) \;=\;a^2\int\cos^2\!\theta\,d\theta

    . . . . . . =\;\frac{a^2}{2}\int(1 + \cos2\theta)\,d\theta \;=\;\frac{a^2}{2}\left(\theta + \tfrac{1}{2}\sin2\theta\right) + C

    . . . . . . =\;\frac{a^2}{2}\left(\theta + \tfrac{1}{2}\cdot2\sin\theta\cos\theta\right) + C \;=\;\frac{a^2}{2}(\theta + \sin\theta\cos\theta) + C .[1]


    Back-substitute: . x \,=\,a\sin\theta \quad\Rightarrow\quad \sin\theta \:=\:\frac{x}{a} \:=\:\frac{opp}{hyp}

    \theta is in a right triangle with: opp = x,\;hyp = a
    Pythagorus gives us: . adj \:=\:\sqrt{a^2-x^2}

    Hence: . \cos\theta \:=\:\frac{\sqrt{a^2-x^2}}{a}

    We have: . \begin{Bmatrix}\theta &=& \sin^{-1}\!\left(\dfrac{x}{a}\right) \\ \\[-3mm] \sin\theta &=& \dfrac{x}{a} \\ \\[-3mm] \cos\theta &=& \dfrac{\sqrt{a^2\!-\!x^2}}{a} \end{Bmatrix}


    Substitute into [1]:

    . . I \;=\;\frac{a^2}{2}\left[\sin^{-1}\!\left(\frac{x}{a}\right) + \frac{x}{a}\cdot\frac{\sqrt{a^2-x^2}}{a}\right] + C

    . . I \;=\;\frac{a^2}{2}\left[\sin^{-1}\!\left(\frac{x}{a}\right) + \frac{x\sqrt{a^2-x^2}}{a^2}\right] + C


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    I have no idea why so many textbooks and sources use \tan^{-1}
    . . which requires extra steps and looks more intimidating.

    \text{Why not use: }\text{sec}^{-1}\left(\frac{a}{\sqrt{a^2-x^2}}\right) \;\text{ and confuse }everyone?

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  8. #8
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    Re: root integral question

    Quote Originally Posted by transgalactic View Post
    how to solve \int\sqrt{a-x^{2}}dx
    i tried by parts but its not not working
    any ideas
    Let x= a sin(t) so that dx= a cos(t) dt. a^2- x^2= a^2- a^2 sin^2(t)= a^2(1- sin^2(t))= a^2 cos^2(t) so that \sqrt{a^2- x^2}= a cos(t)

    \int\sqrt{a^2- x^2}dx= a\int cos^2(t)dt

    And, of course, cos^2(t)= \frac{1}{2}(1+ cos(2t))

    Just realized that this is \int \sqrt{a- x^2}dx not \int\sqrt{a^2- x^2.

    Okay, just replace my "a" with \sqrt{a}: it is
    \sqrt{a}\int cos^2(t)dt= \frac{\sqrt{a}}{2}\int (1+ cos(2t))dt
    = \frac{\sqrt{a}}{2}(t- \frac{1}{2}sin(2t)
    = \frac{\sqrt{a}{2}(sin^{-1}(x/\sqrt{a})- \frac{1}{2}sin(2 sin^{-1}(x/\sqrt{a}})

    sin(2\theta)= 2 sin(\theta)cos(\theta) and cos(sin^{1}(x/\sqrt{a}))= \sqrt{1- x^2/a= \frac{1}{a}\sqrt{a- x^2}
    so the integral is
    \frac{\sqrt{a}}{2}(sin^{-1}(x/\sqrt{a})- \left(\frac{x}{\sqrt{a}}\right)\left(\frac{1}{\sqr  t{a}})(\sqrt{a- x^2}
    = \frac{\sqrt{a}}{2}(sin^{-1}(x/\sqrt{a})- \frac{x}{a\sqrt{a- x^2}})
    Last edited by HallsofIvy; June 25th 2011 at 07:37 AM.
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  9. #9
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    Re: root integral question

    Wow!! Absolutely everyone has mis-read the a as a^2!

    Not massively consequential, but an excuse for some pictures...


    Then, swap the inner function...


    ... where (key in spoiler) ...
    Spoiler:

    ... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case ), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).
    The general drift is...


    Spoiler:



    Spoiler:


    _________________________________________

    Don't integrate - balloontegrate!

    Balloon Calculus; standard integrals, derivatives and methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
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  10. #10
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    Re: root integral question

    i believe the substitution that actually works is u = \sqrt{a}\sin(x)
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