How do you Integrate ( Go pi R^2 ) / ((a^2+x^2)^0.5)? In terms of x?

Question

Where Go = Sigma o, representing surface charge distribution.

Answer 1

(Go pi R^2). ∫dx / ((a^2+x^2)^0.5
Go and R are constant

take x = a tan p dx = a sec^2 p dp

(Go pi R^2) ∫ (a sec^2 p dp ) / a sec p

(Go pi R^2) ∫ sec p dp

=(Go pi R^2) { log [ sec p + tan p]} put back p

x = a tan p so sec^2 p = 1+(x^2/a^2) so
sec p = (a^2+x^2)^1/2/a

=(Go pi R^2) { log [ (x/a) + (a^2+x^2)^1/2/a]}

=(Go pi R^2) { log [ x + (a^2+x^2)^1/2] /a }

=(Go pi R^2) { log [ x + (a^2+x^2)^1/2] - log a }

answer