## anonymous one year ago Integrate with steps shown ((1)/(16-x^2)^(3/2))

1. SolomonZelman

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

2. 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: $\begin{gathered} \int {\frac{{4\cos \theta }}{{64{{\left( {\cos \theta } \right)}^3}}}} d\theta = \hfill \\ \hfill \\ = \frac{1}{{16}}\int {\frac{{d\theta }}{{{{\left( {\cos \theta } \right)}^2}}}} = \frac{1}{{16}}\tan \theta = \frac{1}{{16}}\frac{x}{{\sqrt {16 - {x^2}} }} + k \hfill \\ \end{gathered}$

3. SolomonZelman

oh i was about to post this just now...:) x=sin theta:)

4. SolomonZelman

it is a very nice trig sub.

5. SolomonZelman

but when completing, don't forget to substitute back the x for theta....

6. 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?

7. Michele_Laino

In general, the choice is toward the positive square root, as happens in quantum mechanics

8. 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)

9. SolomonZelman

yeah, just sec²θ.....

10. 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

11. Michele_Laino

or more simply we work with arithmetic radicals only

12. SolomonZelman

I am trying to explore what you said about $$\sqrt{\alpha}$$ a little better... online. Thanks for all of this information:)

13. SolomonZelman

igtg

14. Michele_Laino

:)