Answer:
For the integral ∫▒√(x^2-16 dx), the substitution x = 4 sec θ would be the most helpful.
For the integral ∫▒(x^2 dx)/√(16-x^2 ), the substitution x = 4 sin θ would be the most helpful.
Explanation:
For the integral ∫▒√(x^2-16 dx), we can use the substitution x = 4 sec θ. This is because when we substitute x = 4 sec θ, we get:
dx = 4 sec θ tan θ dθ, and
√(x^2-16) = √(16 sec^2 θ - 16) = 4 tan θ.
Substituting these into the original integral, we get:
∫▒√(x^2-16 dx) = ∫▒4tan^2 θ(4 sec θ tan θ dθ) = 16∫▒tan^3 θ sec θ dθ.
This integral can then be evaluated using techniques such as integration by parts or trigonometric identities.
For the integral ∫▒(x^2 dx)/√(16-x^2 ), we can use the substitution x = 4 sin θ. This is because when we substitute x = 4 sin θ, we get:
dx = 4 cos θ dθ, and
√(16-x^2) = √(16-16sin^2 θ) = 4cos θ.
Substituting these into the original integral, we get:
∫▒(x^2 dx)/√(16-x^2 ) = ∫▒(16sin^2 θ)(4cos θ dθ)/(4cos θ) = 16∫▒sin^2 θ dθ.
This integral can then be evaluated using the half-angle formula for sine or the power-reducing formula for sine.