anonymous
  • anonymous
Integrate with steps shown ((1)/(16-x^2)^(3/2))
Mathematics
jamiebookeater
  • jamiebookeater
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SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \int_{}^{} \frac{1}{\left(16-x^2\right)^{3/2}}~dx}\) it is like this?
Michele_Laino
  • Michele_Laino
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: \[\begin{gathered} \int {\frac{{4\cos \theta }}{{64{{\left( {\cos \theta } \right)}^3}}}} d\theta = \hfill \\ \hfill \\ = \frac{1}{{16}}\int {\frac{{d\theta }}{{{{\left( {\cos \theta } \right)}^2}}}} = \frac{1}{{16}}\tan \theta = \frac{1}{{16}}\frac{x}{{\sqrt {16 - {x^2}} }} + k \hfill \\ \end{gathered} \]
SolomonZelman
  • SolomonZelman
oh i was about to post this just now...:) x=sin theta:)

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SolomonZelman
  • SolomonZelman
it is a very nice trig sub.
SolomonZelman
  • SolomonZelman
but when completing, don't forget to substitute back the x for theta....
SolomonZelman
  • SolomonZelman
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?
Michele_Laino
  • Michele_Laino
In general, the choice is toward the positive square root, as happens in quantum mechanics
SolomonZelman
  • SolomonZelman
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)
SolomonZelman
  • SolomonZelman
yeah, just sec²θ.....
Michele_Laino
  • Michele_Laino
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
Michele_Laino
  • Michele_Laino
or more simply we work with arithmetic radicals only
SolomonZelman
  • SolomonZelman
I am trying to explore what you said about \( \sqrt{\alpha}\) a little better... online. Thanks for all of this information:)
SolomonZelman
  • SolomonZelman
igtg
Michele_Laino
  • Michele_Laino
:)

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