Final answer:
To evaluate the integral ∠ 3/√(9-x²)dx, we use the substitution x = 3 sin(θ) and simplify it to ∠ dθ, which has the solution θ + C. Reverting the substitution gives us sin⁻¹(x/3) + C as our final answer.
Step-by-step explanation:
The integral in question is: ∠ 3/√(9-x²)dx. This integral is reminiscent of the form of the inverse trigonometric function, the inverse sine. We can evaluate it by making a substitution that simplifies the integrand into a form where the inverse trigonometric function can be easily applied.
Let's set x = 3 sin(θ), which implies dx = 3 cos(θ)dθ. Subsequently, our integral becomes ∠ 3/√(9 - (3 sin(θ))²) * 3 cos(θ)dθ. Simplifying inside the square root, 9 - (3 sin(θ))² becomes 9 - 9 sin²(θ) = 9(1 - sin²(θ)), which is 9 cos²(θ) given the Pythagorean identity.
Now the integral is ∠ 3/√(9 cos²(θ)) * 3 cos(θ)dθ. Because √(9 cos²(θ)) is 3 cos(θ), the cosines cancel out and the integral simplifies to ∠ 3/3 dθ, which is just ∠ dθ. The integral of dθ is θ, so we get θ + C for our final answer, where C is the constant of integration.
Lastly, we need to revert our substitution. Since x = 3 sin(θ), sin(θ) = x/3. From this, θ = sin⁻¹(x/3). Therefore, the solution to the original integral is: sin⁻¹(x/3) + C.