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Thread: Integraton by substitution

  1. #1
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    Integraton by substitution

    Hi everyone, just need help with this problem if anyone can point me in the right direction.

    Evaluate \int {0}^{3} s/sqrt(s^2+4^2) ds

    Thanks.
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  2. #2
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    Quote Originally Posted by jo74 View Post
    Hi everyone, just need help with this problem if anyone can point me in the right direction.

    Evaluate \int {0}^{3} s/sqrt(s^2+4^2) ds

    Thanks.
    Is this the correct definite integral?

    \int \frac{dx}{\sqrt{x^2+16}}

    For functions of this form, just use the hyperbolic sine. Set x=sinh (x)
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  3. #3
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    Quote Originally Posted by jo74 View Post
    Hi everyone, just need help with this problem if anyone can point me in the right direction.

    Evaluate \int {0}^{3} s/sqrt(s^2+4^2) ds

    Thanks.
    \int\limits_0^3 {\frac{s}{\sqrt{s^2+16}}\,ds}  = \left\{ \begin{gathered}s^2 = t^2- 16,\,s\,ds = t\,dt \hfill \\<br />
  0 \leqslant s \leqslant 3,\quad 4 \leqslant t \leqslant 5 \hfill \\ <br />
\end{gathered}  \right\} =

    = \int\limits_4^{5} {\frac{t}{{\sqrt {{t^2} - 16 + 16} }}\,dt}  = \int\limits_4^5 {dt} =\, \Bigr. t \Bigr|_4^5 = 5 - 4 = 1.
    Last edited by DeMath; Oct 5th 2009 at 05:16 PM. Reason: typo
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  4. #4
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    Quote Originally Posted by jo74 View Post
    Hi everyone, just need help with this problem if anyone can point me in the right direction.

    Evaluate \int {0}^{3} s/sqrt(s^2+4^2) ds

    Thanks.
    Let u(s) = s^2+4^2 then \dfrac{du}{ds} = 2s~\implies~du = 2s \cdot ds.

    The integral becomes:

    \int \left(\dfrac s{\sqrt{s^2+4^2}} \right)ds = \dfrac12 \int \left(\dfrac {2s}{\sqrt{s^2+4^2}} \right)ds = \dfrac12 \int u^{-\frac12} du

    Can you take it from here?
    Last edited by earboth; Oct 6th 2009 at 02:01 AM. Reason: typo
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