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Math Help - How to simplify this?

  1. #1
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    How to simplify this?

    In my book I have a proof for this integral: \int\frac{dx}{\sqrt{a^2+x^2}}

    \int\frac{dx}{\sqrt{a^2+x^2}} = \int\frac{d\frac{x}{a}}{\sqrt{1+\frac{x^2}{a^2}}} = ln(\frac{x}{a}+\sqrt{1+\frac{x^2}{a^2}})

    {\frac {d}{dx}}ln(\frac{x}{a}+\sqrt{1+\frac{x^2}{a^2}}) = \frac{\frac{x}{a^2\sqrt{\frac{x^2}{a^2}+1}}+\frac{  1}{a}}{\sqrt{\frac{x^2}{a^2}+1}+\frac{x}{a}}

    How can I simplify = \frac{\frac{x}{a^2\sqrt{\frac{x^2}{a^2}+1}}+\frac{  1}{a}}{\sqrt{\frac{x^2}{a^2}+1}+\frac{x}{a}} into \frac{d\frac{x}{a}}{\sqrt{1+\frac{x^2}{a^2}}} ?
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  2. #2
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    You could actually perform the substitution, rather than carrying it in the notation where it seems to be confusing you.

    Do you really mean "proof"? Maybe just a demonstration? The last marvelous step to the natural logarithm is a bit magic.
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  3. #3
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by Bart View Post
    In my book I have a proof for this integral: \int\frac{dx}{\sqrt{a^2+x^2}}

    \int\frac{dx}{\sqrt{a^2+x^2}} = \int\frac{d\frac{x}{a}}{\sqrt{1+\frac{x^2}{a^2}}} = ln(\frac{x}{a}+\sqrt{1+\frac{x^2}{a^2}})

    {\frac {d}{dx}}ln(\frac{x}{a}+\sqrt{1+\frac{x^2}{a^2}}) = \frac{\frac{x}{a^2\sqrt{\frac{x^2}{a^2}+1}}+\frac{  1}{a}}{\sqrt{\frac{x^2}{a^2}+1}+\frac{x}{a}}

    How can I simplify = \frac{\frac{x}{a^2\sqrt{\frac{x^2}{a^2}+1}}+\frac{  1}{a}}{\sqrt{\frac{x^2}{a^2}+1}+\frac{x}{a}} into \frac{d\frac{x}{a}}{\sqrt{1+\frac{x^2}{a^2}}} ?
    That sucks!!! the way it's carried out there.

    This is a prime candidate for tan subsitution.

    \int\frac{dx}{\sqrt{a^2+x^2}}

    Let  x = a tan \theta and  dx = a sec^2 \theta d \theta

    \int\frac{dx}{\sqrt{a^2+x^2}} = \int \frac{ a sec^2 \theta }{ a sec \theta } d \theta = \int sec \theta d \theta = ln| sec \theta + tan \theta | + C

    We arrive at

    ln | \frac{ \sqrt{a^2 + x^2 } }{a} + \frac{x}{a} | + C

    And simplify to

     ln( \sqrt{a^2 + x^2} + x) + C_2

    Noting that  \sqrt{a^2 + x^2} + x > 0  for all x, so we can disregard the absolute value
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  4. #4
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    I don't totally hate what it was trying to do. With a little more experience in the sightly unusual notation, it isn't quite terrible.
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