Integrate ((6-x)/(6+x))^(1/2)?

Question

From 0 to 2

Answer 1

Let u = √(6+x)

u² = 6 + x      => x = u² - 6

2u du = dx
 
∫ √(6−x)/√(6+x) dx

= ∫ √(6−(u²−6))/u * 2u du

= ∫ 2√(12−u²) du 

u² = 12sin²t

u = 2√3 sint

du = 2√3 cost dt

= ∫ 2√(12−12sin²t) * 2√3 cost dt

= ∫ 2 * 2√3 cost * 2√3 cost dt

= ∫ 24 cos²t dt

= ∫ (12 + 12 cos(2t)) dt

= 12t + 6 sin(2t) + C

= 12t + 12 sint cost + C

= 12 sin⁻¹(u/(2√3)) + 12 u/(2√3) √(12−u²)/(2√3) + C

= 12 sin⁻¹(u/(2√3)) + u √(12−u²) + C

= 12 sin⁻¹(√(6+x)/(2√3)) + √(6+x) √(12−(6+x)) + C

= 12 sin⁻¹(√(6+x)/(2√3)) + √(6+x) √(6−x) + C

From 0 to 2:

= 12 sin⁻¹(√8/(2√3)) + √8√4 − 12 sin⁻¹(√6/(2√3)) − √6√6

= 4√2 − 6 + 12sin⁻¹(√(2/3)) − 12sin⁻¹(1/√2)

= 4√2 − 6 + 12sin⁻¹(√(2/3)) − 12 * π/4

= 4√2 − 6 − 3π + 12sin⁻¹(√(2/3))

= 1.695875706

∫ √(6−x)/√(6+x) dx = 1.695875706