Question

integral of x/(25 - x^2)^1/2

Answer 1

\[\int\limits_{?}^{?} x/\sqrt{25 - x^{2}}\]

Answer 2

using substitution: -sqrt(25-x^2) + C, C a constant

Answer 3

use trig sub like last time = you have the same thing only backwards

Answer 4

use the substitution x=asin(x)

Answer 5

could i do x = 5cos(theta)? then dx = -5sin(theta)?

Answer 6

x=5sin(theta)

Answer 7

there is a special trig relationship i'm not sure what it is you'd have to ask jim thompson or someone who really knows his stuff. it carries throughout calc as you can see it in inverses like i said before

Answer 8

let u = 25 - x^2 then du = -2x
so integral -1/2 du/u^(1/2) = - u^(1/2)
substituting back: Ans -sqrt(25-x^2) + constant

Answer 9

give me a sec i'll attach a pic