Question

Integrate with steps shown

((1)/(16-x^2)^(3/2))

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

it is a very nice trig sub.

Answer 5

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

Answer 6

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?

Answer 7

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

Answer 8

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)

Answer 9

yeah, just sec²θ.....

Answer 10

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

Answer 11

or more simply we work with arithmetic radicals only

Answer 12

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

Answer 13

igtg

Answer 14

:)