Let I = ∫√(a² - x² )dx ............1

Put x = a*sinθ

=> dx = a*cosθ * dθ

From equation 1, we get

      I = ∫√(a² - (a*sinθ)² )*a*cosθ * dθ

=> I = ∫√(a² - a² *sin² θ )*a*cosθ * dθ

=> I = a² *∫√(1 -  *sin² θ )*cosθ * dθ

=> I = a² *∫√(cos² θ)*cosθ* dθ

=> I = a² *∫cos θ*cosθ* dθ

=> I = a² *∫cos² θ dθ

=> I = a² *∫(1 + cos 2θ)/2 dθ

=> I = (a² /2)*∫(1 + cos 2θ) dθ

=> I = (a² /2)*[θ + (sin 2θ)/2] + C ..........2

Now x = a*sinθ

=> sinθ = x/a

and cosθ = √(a² - x² )/a {from Pythagorus theorem} 

=> θ = sin^-1 (x/a)

From eqaution 2, we get

      I = (a² /2)*[sin^-1 (x/a) + {sin 2θ}/2] + C

=> I = (a² /2)*[sin^-1 (x/a) + {2*sin θ*cos θ}/2] + C

=> I = (a² /2)*[sin^-1 (x/a) + {(x/a)*(√(a² - x² )/a)}] + C

=> I = (x/2)*(√(a² - x² ) + (a² /2)*{sin^-1 (x/a)} + C

So,  ∫√(a² - x² )dx = (x/2)*(√(a² - x² ) + (a² /2)*{sin^-1 (x/a)} + C