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Math Help - How To integrate this?

  1. #1
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    How To integrate this?

    <br />
\int (x^2)/(\sqrt(9+x^2)) dx<br />

    ** Thats the best I can make this equation look...

    Anyways I have been trying to get this thing for hours, I clearly do not know what I'm doing. Advice geatly apriciated!
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  2. #2
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    I don't know if it works out or not but try x = 3tany
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  3. #3
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    Yea i have tried that over and over but Im getting no where
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  4. #4
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    after the trig sub, do by parts
    u = tany
    dv = secytany

    then to integrate sec^3y you gotta do by parts again
    good luck
    Last edited by mr fantastic; March 2nd 2009 at 02:51 AM. Reason: Merged posts
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  5. #5
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    Okay sooo I do all that and then I get to the part where it loops back to my original integral. But since i use integration by parts twice the two negatives make a positive and then you have somthing like:

    A = original integral
    A = stuff + A
    so i clearly did somthing wrong...

    Heres my step my step:
    <br />
\int (x^2)/(\sqrt(9-x^2)) = 9\int (tan(x)^2sec(x) dx<br />
    <br />
9\int (tan(x)^2sec(x) dx = 9[tan(x)sec(x) - \int sec(x)^3<br />
    <br />
9[tan(x)sec(x) - \int sec(x)^3 = 9[tan(x)sec(x) - [sec(x)tan(x) - \int tan(x)^2sec(x)]]<br />
    <br />
9\int (tan(x)^2sec(x) dx = 9tan(x)sec(x) - 9sec(x)tan(x) + 9\int (tan(x)^2sec(x) dx<br />

    At this point its clearly wrong what did I do?
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  6. #6
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    integrate[x^2/sqrt.(9-x^2)] dx

    now,x=3cosA.

    integrate(9 cos^2 A/sqrt.(9-9cos^2 A))*-3 sinA da

    =-integrate{9 cos^2 A * dA}

    =-9/2 integrate{(1+cos 2A) dA}

    the rest is simple!!!!!!!!!!
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  7. #7
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    ooops the question is:

    x^2/ sqrt(9+x^2)

    Some reason I put a minus before, long day...

    So tan sub needs to be used
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  8. #8
    Super Member redsoxfan325's Avatar
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    \int \frac{x^2}{\sqrt{9+x^2}}\,dx

    Integration by parts:

    Let x=u and dv=\frac{x}{\sqrt{9+x^2}}

    Thus, dx=du and v=\sqrt{9+x^2}

    Your integral is now: x\sqrt{9+x^2}-\int \sqrt{9+x^2}\,dx

    Use x=3tan(t) and dx=3sec^2t.

    Now you have: x\sqrt{9+x^2}-\int \sqrt{9+9tan^2t}*3sec^2t\,dt = x\sqrt{9+x^2}-9\int sec^3t\,dt

    Let S=\int sec^3t\,dt

    Integration by parts again:

    Let u=sec(t) and dv=sec^2(t).

    Thus du=sec(t)tan(t) and v=tan(t) and the integral becomes: sec(t)tan(t)-\int tan^2(t)sec(t)\,dt.

    This becomes: sec(t)tan(t)-\int (sec^2(t)-1)sec(t)\,dt = sec(t)tan(t)-\int sec^3(t) + \int sec(t)\,dt.

    Thus S = sec(t)tan(t) - S + \int sec(t)\,dt and 2S = sec(t)tan(t) + \int sec(t)\,dt.

    Using a table of integrals, we find that \int sec(t)\,dt = ln|sec(t)+tan(t)|.

    Thus 2S = sec(t)tan(t) + ln|sec(t)+tan(t)|.

    Hence, \int sec^3t\,dt = S = \frac{1}{2}sec(t)tan(t) + \frac{1}{2}ln|sec(t)+tan(t)|.

    Plugging \int sec^3t\,dt back into the equation a few lines up, we have x\sqrt{9+x^2}-9(\frac{1}{2}sec(t)tan(t) + \frac{1}{2}ln|sec(t)+tan(t)|).

    Remember though, that since x = 3tan(t), t = tan^{-1}\frac{x}{3}.

    Thus, x\sqrt{9+x^2}-9(\frac{1}{2}sec(t)tan(t) + \frac{1}{2}ln|sec(t)+tan(t)|) becomes x\sqrt{9+x^2}-9(\frac{1}{2}sec(tan^{-1}\frac{x}{3})tan(tan^{-1}\frac{x}{3}) + \frac{1}{2}ln|sec(tan^{-1}\frac{x}{3})+tan(tan^{-1}\frac{x}{3})|).

    Using the triangle, if the tangent of an angle is \frac{x}{3}, the secant of that angle is \frac{3}{\sqrt{9+x^2}}. Plugging everything makes that nasty equation become: x\sqrt{9+x^2}-9(\frac{1}{2}*\frac{3}{\sqrt{9+x^2}}*\frac{x}{3}) + \frac{1}{2}ln|\frac{3}{\sqrt{9+x^2}}+\frac{x}{3}|).

    Simplifying that yields: x\sqrt{9+x^2}-\frac{9}{2}*\frac{x}{\sqrt{9+x^2}} + \frac{9}{2}ln|\frac{3}{\sqrt{9+x^2}}+\frac{x}{3}|, which is what I think the answer is.

    (You think that's ugly, you should see what it looks like in LaTex.)
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  9. #9
    Super Member redsoxfan325's Avatar
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    I just realized I made a mistake at the end. When I said sec(arctan(\frac{x}{3}))=\frac{3}{\sqrt{9+x^2}}, I was wrong. It's actually \frac{\sqrt{9+x^2}}{3}, making the final answer:

    \frac{1}{2}x\sqrt{9+x^2}-\frac{9}{2}ln(\frac{1}{3}\sqrt{9+x^2}+\frac{1}{3}x  )

    Sorry for the mistake.
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  10. #10
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    Haha no worrys, as it turns out there is a formula in the back of my text book that goes directly from the integral to exactly what you came up with :P I really appriciate the help thank you
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