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Math Help - Trig Substitution with Completing the Square

  1. #1
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    Trig Substitution with Completing the Square

    Integral of:

    dx/(sqrt(x^2+2x+65)

    I completed the square on the denominator to rewrite it as:

    sqrt((x+1)^2+64)

    When I try to solve this, I let:

    u=x+1
    du=dx

    So we get the integral of:

    du/sqrt(u^2+64)

    Then I set:
    u=8tan(theta)
    du=8sec^2(theta)d(theta)

    This simplifies to just the integral of:

    sec(theta)d(theta)

    The integral of this is:

    ln(abs(tan(theta)sec(theta)) + C

    When I setup a right triangle, I get the following sides:

    Hypotenuse:sqrt((x+1)^2+64)
    Adjacenet: 8
    Opposite: x+1

    When I use these values to replace tan(theta) and sec(theta) my final answer is:

    ln(abs((x+1)/8) * (sqrt((x+1)^2+64)/8)) + C

    However, the online website has a different answer. It has the same basic terms but without dividing by 8. Where am I going wrong?
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Trig Substitution with Completing the Square

    The two forms are equivalent, as you may write:

    \ln\left(\frac{f(x)}{8} \right)+C=\ln(f(x))-\ln(8)+C=\ln(f(x))+C

    Since \ln(8) in a constant, it may be combined with the constant of integration.
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  3. #3
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    Re: Trig Substitution with Completing the Square

    Hello, vesperka!

    Your final line should have a Plus (+).



    \ln\left|\frac{x+1}{8} + \frac{\sqrt{x^2+2x +65}}{8}\right| + C

    However, the online website has a different answer.
    It has the same basic terms but without dividing by 8.
    Where am I going wrong? . Your answer is correct!

    They did some "invisible" (very sneaky) simplifying.

    You had: . \ln\left|\frac{x+1}{8} + \frac{\sqrt{x^2+2x +65}}{8}\right| +\;c

    . . . . . . =\;\ln\left|\frac{x+1 + \sqrt{x^2+2x+65}}{8}\right| +\; c

    . . . . . . =\;\ln\left|x + 1 + \sqrt{x^2 + 2x + 65}\right|\; - \underbrace{\ln(8)\;+\;c}_{\text{This is a constant}}

    . . . . . . =\;\ln\left|x + 1 + \sqrt{x^2+2x+65}\right| + C


    Yes . . . just one more thing to watch for.
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  4. #4
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    Re: Trig Substitution with Completing the Square

    Thanks for the replies guys. I didn't realize this, but for the integral of sec(theta) I was multiplying the tangent and secant inside of the natural log when they actually should be added together. Thanks for the explanation on why -ln(8) is a constant that can be factored out!
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