A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
Integrate with steps shown
((1)/(16x^2)^(3/2))
anonymous
 one year ago
Integrate with steps shown ((1)/(16x^2)^(3/2))

This Question is Closed

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{black}{ \displaystyle \int_{}^{} \frac{1}{\left(16x^2\right)^{3/2}}~dx}\) it is like this?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2if 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
 one year ago
Best ResponseYou've already chosen the best response.1oh i was about to post this just now...:) x=sin theta:)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1it is a very nice trig sub.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1but when completing, don't forget to substitute back the x for theta....

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1i 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
 one year ago
Best ResponseYou've already chosen the best response.2In general, the choice is toward the positive square root, as happens in quantum mechanics

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1oh, 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
 one year ago
Best ResponseYou've already chosen the best response.1yeah, just sec²θ.....

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2yes! 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
 one year ago
Best ResponseYou've already chosen the best response.2or more simply we work with arithmetic radicals only

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1I am trying to explore what you said about \( \sqrt{\alpha}\) a little better... online. Thanks for all of this information:)
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.