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Thread: trig substitution problem

  1. #1
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    trig substitution problem

    x/sqrt(9-x)

    u = x
    a = 3

    i used.... x=asinx

    i end up substituting this into the problem....
    x = theta

    (3cosx)/3cosx

    i then reduce it to

    3cosx and integrate it to get....
    3sinx + c

    now i use my triangle and solve in terms of x and now i get as a final awnser...

    x + c

    this doesnt seem right, can anyone verify this for me!
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  2. #2
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    It can't be right. The derivative of $\displaystyle x + C$ is not $\displaystyle \dfrac{x^2}{\sqrt{9-x^2}}$.

    After performing the trigonometric substitution, the numerator should be $\displaystyle \left ( 3\sin{\theta} \right ) ^2$, not $\displaystyle \left ( 3\cos{\theta}{ \right ) ^2$.
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  3. #3
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    ok

    (9sinx / 3cosx) dx

    i integrated this properly i think and worked it to 9/3(ln|secx+tanx|-sinx) + C can anyone verify this for me?
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  4. #4
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    $\displaystyle \displaystyle \int \frac{x^2}{\sqrt{9-x^2}} \, dx$

    $\displaystyle x = 3\sin{t}$

    $\displaystyle dx = 3\cos{t} \, dt$

    $\displaystyle \displaystyle \int \frac{9\sin^2{t}}{\sqrt{9 - 9\sin^2{t}}} \cdot 3\cos{t} \, dt$

    $\displaystyle \displaystyle \int \frac{9\sin^2{t}}{3\sqrt{1 - \sin^2{t}}} \cdot 3\cos{t} \, dt$

    $\displaystyle \displaystyle \int 9\sin^2{t} \, dt$

    $\displaystyle \displaystyle \frac{9}{2} \int 1 - \cos(2t) \, dt$

    $\displaystyle \displaystyle \frac{9}{2} \left[t - \frac{\sin(2t)}{2}\right] + C$

    $\displaystyle \displaystyle \frac{9}{4} \left[2t - \sin(2t)\right] + C$

    $\displaystyle \displaystyle t = \arcsin\left(\frac{x}{3}\right)$

    $\displaystyle \displaystyle \sin(2t) = 2\sin{t}\cos{t} = \frac{2x}{3} \cdot \sqrt{1 - \left(\frac{x}{3}\right)^2} = \frac{2x}{9} \sqrt{9 - x^2}$

    so, antiderivative is ...

    $\displaystyle \displaystyle \frac{9}{2}\left[\arcsin\left(\frac{x}{3}\right) - \frac{x \sqrt{9-x^2}}{9}\right] + C$
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