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Math Help - Integral 07

  1. #1
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    Integral 07

    \int \frac{sec(\sqrt{x})dx}{\sqrt{x}}
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  2. #2
    Super Member wingless's Avatar
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    Don't you have any thought on this one?
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  3. #3
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    Integration by parties?

    u=\sqrt{x}
    v=\int sec(\sqrt{x})dx
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  4. #4
    Super Member wingless's Avatar
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    No, u substitution. Let u = \sqrt x ...
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  5. #5
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    but in denominator is sec(\sqrt{x})
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  6. #6
    Super Member wingless's Avatar
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    Just do what I said and you'll see why I did it..

    Let u = \sqrt x... Now what's du ?
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  7. #7
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    du=\frac{2x^\frac{3}{2}}{3}
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  8. #8
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    \int \frac{sec(\sqrt{x})dx}{\sqrt{x}}=\int {2\sec (u)du}\;\;,\; u=\sqrt{x}
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  9. #9
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    Why?
    =\int 2sec(u)du
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  10. #10
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    Surely you see it.

    If we let u=\sqrt{x}, then du=\frac{1}{2\sqrt{x}}dx

    and 2du=\frac{1}{\sqrt{x}}dx

    See?. The 2du replaces the \frac{1}{\sqrt{x}}dx in your integral.

    Then, we get 2\int sec(u)du
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  11. #11
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Apprentice123 View Post
    Why?
    =\int 2sec(u)du
    You make the substitution u=\sqrt{x}\implies\,du=\frac{\,dx}{2\sqrt{x}}

    \therefore\int\frac{\sec(\sqrt{x})}{\sqrt{x}}\,dx\  implies 2\int\frac{\sec(u)}{2\sqrt{x}}\,dx\implies 2\int\sec(u)\,du

    --Chris
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  12. #12
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    What kind of integral you used?
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