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Math Help - trig substitution

  1. #1
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    trig substitution

    1. (integral) dx / sqrt(9+x^2)
    I had a^2 = 9, a=3, x = 3tan(theta), dx = 3(sec(theta))^2, and
    sqrt(9+x^2) = 3sec(theta)

    I plugged them in and got:

    3(sec(theta))^2 / 3(sec(theta)) = (integral) 3sec(theta).

    I then integrated it and got ln|sec(theta) + tan(theta)| +c

    I just divided the 3 and got: sqrt(9+x^2) / 3

    and for tan(theta), i divided the 3 again and got tan(theta)=x/3

    So, the ending result I had gotten was :

    ln|(sqrt(9+x^2) / 3) + (x/3)| +c But the answer in the book had the same answer, but the 3's were not in. What am I doing wrong??

    2.(3/2,0)(integral) dx/sqrt(9- x^2)

    I used x = 3sin(theta), a = 3, dx = 3cos(theta), and

    sqrt(9- x^2) = 3cos(theta)

    I got the integral: 3cos(theta)/3cos(theta) = integral 1 (theta)

    I then integrated it and just got theta.

    I inversed the sin and got x = 3sin(theta)^-1

    My problem is that I plugged the numbers in, and it was undefined. the

    answer in the book came out to pi/6

    plz help
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  2. #2
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    1) With sqrt(1 + x^2) in the denominator you'll find sinh^-1 an easier sub.

    Just in case a picture helps...



    ... where



    ... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to x, 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).

    Spoiler:



    Ok, your way is equally good, and to answer your question, actually...

    ln|(sqrt(9+x^2) / 3) + (x/3)|

    = ln|(sqrt(9+x^2 + x) / 3)|

    = ln|sqrt(9+x^2 + x)| - ln|3|

    and ln|3| can be part of the constant.

    Anyway, hope you were not too unpleasantly diverted by the pics.


    __________________________________________
    Don't integrate - balloontegrate!
    Balloon Calculus; standard integrals, derivatives and methods
    Balloon Calculus Drawing with LaTeX and Asymptote!
    Last edited by tom@ballooncalculus; February 5th 2010 at 01:48 PM.
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  3. #3
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    I am sorry, I do not understand the graph, this is for Calc II class
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  4. #4
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    Pardon the diversion, anyway I edited in the relevant point for no. 1 above.

    2). You aren't clear whether you plugged in values for x or theta, probably x, and maybe your inversing is muddled - integral should be

    sin^-1(x/3)

    Edit:

    Again, just in case it helps...

    Spoiler:
    Last edited by tom@ballooncalculus; February 5th 2010 at 02:14 PM.
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  5. #5
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    Oh!! I got it now for #1. Thank you!
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  6. #6
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    Hello, driver327!

    2)\;\;\int^{\frac{3}{2}}_0\frac{dx}{\sqrt{9- x^2}}
    This is a standard form

    . . There is a formula: . \int\frac{du}{\sqrt{a^2-u^2}} \;=\;\arcsin\left(\frac{u}{a}\right) + C


    If you want to do it out:

    . . \text{Let: }\:x \,=\,3\sin\theta \quad\Rightarrow\quad dx \,=\,3\cos\theta\,d\theta \quad\Rightarrow\quad \sqrt{9-x^2} \,=\,3\cos\theta


    Substitute: . \int\frac{3\cos\theta\,d\theta}{3\cos\theta} \;=\;\int d\theta \;=\;\theta + C


    Back-substitute: . 3\sin\theta = x \quad\Rightarrow\quad \sin\theta \:=\:\frac{x}{3} \quad\Rightarrow\quad \theta \:=\:\arcsin\frac{x}{3}
    . . So we have: . \arcsin\frac{x}{3} \,\bigg]^{\frac{3}{2}}_0


    Evaluate: . \arcsin\left(\frac{1}{3}\cdot\frac{3}{2}\right) - \arcsin\left(\frac{0}{3}\right) \;\;=\;\;\arcsin\frac{1}{2} - \arcsin 0 \;\;=\;\;\frac{\pi}{6} - 0 \;\;=\;\;\frac{\pi}{6}

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