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anonymous

  • one year ago

Integrate with steps shown ((1)/(16-x^2)^(3/2))

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  1. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \int_{}^{} \frac{1}{\left(16-x^2\right)^{3/2}}~dx}\) it is like this?

  2. Michele_Laino
    • one year ago
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    if we make this substitution: x=4*sin(\theta), we can rewrite our integral as follows: \[\int {\frac{{4\cos \theta }}{{64{{\left( {\cos \theta } \right)}^3}}}} d\theta \] so we have the subsequent steps: \[\begin{gathered} \int {\frac{{4\cos \theta }}{{64{{\left( {\cos \theta } \right)}^3}}}} d\theta = \hfill \\ \hfill \\ = \frac{1}{{16}}\int {\frac{{d\theta }}{{{{\left( {\cos \theta } \right)}^2}}}} = \frac{1}{{16}}\tan \theta = \frac{1}{{16}}\frac{x}{{\sqrt {16 - {x^2}} }} + k \hfill \\ \end{gathered} \]

  3. SolomonZelman
    • one year ago
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    oh i was about to post this just now...:) x=sin theta:)

  4. SolomonZelman
    • one year ago
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    it is a very nice trig sub.

  5. SolomonZelman
    • one year ago
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    but when completing, don't forget to substitute back the x for theta....

  6. SolomonZelman
    • one year ago
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    i see a little gap there \(\large\color{black}{ \displaystyle \sqrt{F^2}=|F|}\) so, shouldn't it be absolute value of cos(x) technically?

  7. Michele_Laino
    • one year ago
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    In general, the choice is toward the positive square root, as happens in quantum mechanics

  8. SolomonZelman
    • one year ago
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    oh, nvm, i guess in this case it is just that the powers cancels.... I am overthinking myself. Absolute value is not there.... (quantum mechanics? :D)

  9. SolomonZelman
    • one year ago
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    yeah, just sec²θ.....

  10. Michele_Laino
    • one year ago
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    yes! In quantum mechanics it is convention to consider the positive square root of operator \alpha Namely if \alpha is an operator, then +sqrt(\alpha) is the square root operator

  11. Michele_Laino
    • one year ago
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    or more simply we work with arithmetic radicals only

  12. SolomonZelman
    • one year ago
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    I am trying to explore what you said about \( \sqrt{\alpha}\) a little better... online. Thanks for all of this information:)

  13. SolomonZelman
    • one year ago
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    igtg

  14. Michele_Laino
    • one year ago
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    :)

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