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Math Help - Integration? Hyperbolic?

  1. #1
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    Integration? Hyperbolic?

    <br />
\int_{0}^{2}  \frac{1}{\sqrt{256 +x^2}} dx<br />

    Any suggestion on how to truck though this?
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  2. #2
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    Quote Originally Posted by Lolcats View Post
    <br />
\int_{0}^{2} \frac{1}{\sqrt{256 +x^2}} dx<br />

    Any suggestion on how to truck though this?
    Try the substitution x=16tan(\theta).
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  3. #3
    Member Nacho's Avatar
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    \tan \theta=\dfrac{x}{16}
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  4. #4
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    Try the substitution .
    Whats the reasoning behind this?
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  5. #5
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    Oh nvm thanks Nacho!
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  6. #6
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    k so I have <br />
\displaystyle{\frac{1}{\sqrt{256+16\tan\!\left(u^{  2}\right)}}\frac{du}{16\sec^{2}\!\left(u\right)}}<br />
    where do I go from that?

    Sorry if i dont respond right away, my laptops about to die and I have a chemistry seminar for the next hour just been trying to get some math done while I wait
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  7. #7
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    You should get

    \int \frac{\sec^2 u * du}{\sqrt{256 + 256 \tan^2 u}}

    after the substitution.
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  8. #8
    Member Nacho's Avatar
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    Quote Originally Posted by Lolcats View Post
    k so I have <br />
\displaystyle{\frac{1}{\sqrt{256+16\tan\!\left(u^{  2}\right)}}\frac{du}{16\sec^{2}\!\left(u\right)}}<br />
    where do I go from that?
    Be careful! <br />
\left( {16\tan \theta } \right)^2  \ne 16\tan \left( {\theta ^2 } \right)<br />

    And you donīt forget that: 1+\tan^2 \theta=\sec^2 \theta
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  9. #9
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    You should get



    after the substitution.
    Isn't it

    <br />
\displaystyle{\frac{1}{\sqrt{256+256\tan^{2}\!\lef  t(u\right)}}\!\left(16\sec^{2}\!\left(u\right)\rig  ht)du}<br />

    after the substitution?
    Cuz the derivative of 16tan(u) = (16sec(u))^2
    ?
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  10. #10
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    Now factor the denominator and use the identity 1+ \tan^2 \theta = \sec^2 \theta
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  11. #11
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    256=16^2 is the reasoning(correct me if i am wrong!).
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  12. #12
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    k alright, after that I have...

    <br />
\displaystyle{\frac{\sec^{2}\!\left(u\right)du}{\s  qrt{\sec^{2}\!\left(u\right)}}}<br />
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  13. #13
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    i guess can you just sqrt the sec(U)^2? that would make this pretty easy then
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  14. #14
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    <br />
\displaystyle{\ln\!\left(\left|\tan\!\left(u\right  )+\sec\!\left(u\right)\right|\right)}<br />
    And now I'm utterly lost how to sub back u in... I would assume u=tan^(-1)(x)/(16)... but what do I do with something like that...?

    This question looked so harmless to begin with...
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  15. #15
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Lolcats View Post
    <br />
\int_{0}^{2}  \frac{1}{\sqrt{256 +x^2}} dx<br />

    Any suggestion on how to truck though this?
    Alternatively, try x=16\sinh(x)
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