Question

Integral da raiz de [9 - (x-1)^2]

Answer 1

Olá,

∫ √9-(x-1)² = ∫ √9-u² du = 3 ∫ √9-9 sen²(s) cos(s) ds

u = (x-1)
u = 3 sen(s) ⇒ sen(s) = u/3 ⇒ s = arc sen(u/3)
u² = 9 sen²(s)
du = 3 cos(s) ds

3 ∫ √9-9 sen²(s) cos(s) ds = 3 ∫ √3²(1-sen²(s)) cos(s) ds
9 ∫ √1-sen²(s) cos(s) ds = 9 ∫ √cos²(s) cos(s) ds
9 ∫ cos(s) cos(s) ds = 9 ∫ cos²(s) ds = 9 (s/2 + cos(s)sen(s)/2)

sen(s) = u/3
cos(s) = √1-sen²(s) = √1-(u/3)² = √1-u²/9 = (√9-u²)/3

Substituindo:

9/2 (s + cos(s)sen(s) = 9/2 (arc sen(u/3) + (√9-u²)/3(u/3)) 
9/2 (arc sen(u/3) + (u√9-u²)/9)
9/2 arc sen(u/3) + 9/9 (u√9-u²)/2
9/2 arc sen(u/3) + (u√9-u²)/2

Substituindo o valor de u = (x-1) teremos a resposta:

9/2 arc sen((x-1)/3) + ((x-1)√9-(x-1)²)/2 + c