anonymous
  • anonymous
Integrate with steps shown ((1)/(16-x^2)^(3/2))
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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:)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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
:)

Looking for something else?

Not the answer you are looking for? Search for more explanations.