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Math Help - Integration

  1. #1
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    Integration

    Is there a way to do this from the second expression? I ended up where I started from. I do not want to use trigonometrical substitution.
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  2. #2
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    Re: Integration

    No easy way.

    Just stick with trig. sub. ... or (better yet) use hyperbolic substitution.

    sec^2(x) - 1 =tan^2(x) . . . . ... . . . cosh^2(x) - 1 = sinh^2(x) .
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  3. #3
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    Re: Integration

    This is what the question says to do. It must be possible to use the second expression.
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  4. #4
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    Re: Integration

    Quote Originally Posted by Stuck Man View Post
    This is what the question says to do. It must be possible to use the second expression.
    Your Original Post gave the impression that you did not want to use trig substitution.
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  5. #5
    Member sbhatnagar's Avatar
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    Angry Re: Integration

    There is one way to solve it without trig substitutiton but it does not involve the second expression.

    \int_{5/4}^{5/3}\frac{du}{\sqrt{u^2-1}}

    =\int_{5/4}^{5/3}\frac{(u+\sqrt{u^2-1})}{(u+\sqrt{u^2-1})\sqrt{u^2-1}}du

    =\int_{5/4}^{5/3}\frac{\frac{u}{\sqrt{u^2-1}}+1}{(u+\sqrt{u^2-1})}du

    (This is \frac{f'(u)}{f(u)} form)

    =[\ln{(u+\sqrt{u^2-1})}]_{5/4}^{5/3}

    Note- The identity could also have been used : \int \frac{1}{\sqrt{x^2-a^2}}dx=\ln{(x+\sqrt{x^2-a^2})}+C
    Last edited by sbhatnagar; October 17th 2011 at 02:11 AM.
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  6. #6
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    Re: Integration

    I've not seen it before. The question says to use the the two fractions.
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  7. #7
    Member sbhatnagar's Avatar
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    Re: Integration

    Look, stuck man, I donot think there is any way to find the integral from the 2 fractions. Even wolfram alpha solves it using trigonometric substitution. You can see it here if you like. link. If required you can also use hyperbolic substitution.
    Last edited by sbhatnagar; October 17th 2011 at 03:08 AM.
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