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Math Help - Another integration question!

  1. #1
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    Another integration question!

    Question: Using the substitution x = 1/y, or otherwise, find: (see attached)


    Sorry for the picture quality, I took it off my phone.

    It's x and in the bracket, x^2 you see there.

    Thanks in advance!
    Attached Thumbnails Attached Thumbnails Another integration question!-image003.jpg  
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  2. #2
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    Quote Originally Posted by margaritas
    Question: Using the substitution x = 1/y, or otherwise, find: (see attached)

    Sorry for the picture quality, I took it off my phone.
    It's x and in the bracket, x^2 you see there.
    Thanks in advance!
    Hello, margaritas,

    do you mean: \int{\frac{1}{x \cdot \sqrt{x^2-1}}dx}

    If so the result is \arctan \left( \sqrt{x^2-1} \right)

    Steps to reach this result will follow.

    Greetings

    EB
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  3. #3
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    Thanks for the help, EB, yep that's the question. But the answer as given by my tutor is supposedly: C - sin^(-1) (1/x).
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  4. #4
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    Quote Originally Posted by margaritas
    Question: Using the substitution x = 1/y, or otherwise, find: (see attached)...
    Hello, margaritas,
    it's me again.

    As you suggested, use x=\frac{1}{y}\ \Longrightarrow\ \frac{dx}{dy}=-\frac{1}{y^2}. So {dx}=-\frac{1}{y^2}\cdot {dy}.

    Substitute the x and the dx and you'll get:

    \int{\frac{1}{\frac{1}{y}\cdot \sqrt{\frac{1}{y^2}-1}}\cdot \left(-\frac{1}{y^2}\right)\cdot dy}. Rearrange:

    \int{\frac{y\cdot y}{\sqrt{1-y^2}}\cdot \left(-\frac{1}{y^2}\right) \cdot dy} = -\int{\frac{1}{\sqrt{1-y^2}} \cdot dy}

    As you may see, this is -\int{\frac{1}{\sqrt{1-y^2}} \cdot dy} = -\arcsin(y)+C=C-\arcsin\left(\frac{1}{x}\right)

    So the answer given in your book is OK - but my result is true too. There are integrals with different solutions. After a few steps of transformation, you'll reach my result. Have a look here:
    http://www.mathhelpforum.com/math-he...ead.php?t=4278

    Greetings

    EB
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  5. #5
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    Great work, EB!

    It seems to invite Trig Substitution,
    . . so I tried it . . . and found yet another solution.

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

    Let x = \sec\theta\quad\Rightarrow\quad dx = \sec\theta\tan\theta\,d\theta\quad\Rightarrow\quad  \sqrt{x^2 - 1} = \tan\theta

    Substitute: . \int\frac{\sec\theta\tan\theta\,d\theta}{\sec \theta\tan\theta}\;=\;\int d\theta \;=\;\theta + C

    Back-substitute: . \boxed{\text{arcsec }x + C}

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

    Our three answers are equivalent.

    I had: \sec\theta = \frac{x}{1}

    \theta is an angle is a right triangle with: hyp = x,\;adj = 1
    Code:
                      *
              x    *  |  ______
                *     | √x - 1
             * θ      |
          * - - - - - * 
                1

    Your answer . \arctan\sqrt{x^2-1} .is the same angle.


    The other acute angle is: \frac{\pi}{2} - \theta . and \sin\left(\frac{\pi}{2} - \theta\right) = \frac{1}{x}

    Hence: . \frac{\pi}{2} - \theta\:=\:\arcsin\left(\frac{1}{x}\right)\quad \Rightarrow\quad\theta \:= \:\frac{\pi}{2} - \arcsin\left(\frac{1}{x}\right)

    . . which verifies margaritas' solution.

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  6. #6
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    Thanks for the help, guys!
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