How to integrate

Question

sqrt (16-x^2)

Answer 1

The expression is ʃ √(16 - x²) dx.

= ʃ √(4² - x²) dx

Let x = 4 sin(θ) ⇒ dx = 4 cos(θ) dθ and √(4² - x²) = 4 cos(θ).

= ʃ 4 cos(θ) * 4 cos(θ) dθ

= 16 ʃ cos²(θ) dθ

= 16 ʃ [1+cos(2θ)]/2 dθ                                 [ Since cos²(θ) = [1+cos(2θ)]/2 ]

= 8 [θ + sin(2θ)/2 ] + C

= 8 [θ + sin(θ) * cos(θ) ] + C                       [ Since sin(2θ) = 2 sin(θ) * cos(θ) ]

= 8 [θ + sin(θ) * √(1 - sin² θ) ] + C              [ Since cos(θ) = √(1 - sin² θ) ]

= 8 [sin^-1(x/4) + (x/4)* √(1 - x²/16) ] + C   [ Since x = 4 sin(θ) ]

= (1/2) [16 sin^-1(x/4) + x√(4² - x²) ] + C

=8 sin^-1(x/4) + (x/2)√(16 - x²) ] + C

ʃ √(16 - x²) dx = 8 sin^-1(x/4) + (x/2)√(16 - x²) ] + C