Final Answer:
The integral ∫ x²√(3+2x-x²) dx can be evaluated using trigonometric substitution as follows: let x = √3 sinθ + 1. The final result is (1/6) [(√3 sinθ + 1)³ + C], where C is the constant of integration.
Step-by-step explanation:
To evaluate the given integral, the method of trigonometric substitution can be employed. Initially, factorize the expression under the square root: 3 + 2x - x² = (x - 1)². To simplify further, x - 1 = ±√(3 + 2x - x²).
Let's set x - 1 = √(3 + 2x - x²), then x = √(3 + 2x - x²) + 1. To eliminate the square root, substitute x = √3 sinθ + 1. Thus, dx = √3 cosθ dθ.
Now substitute these values into the integral: ∫ x²√(3+2x-x²) dx becomes ∫(√3 sinθ + 1)² * √(3 + 2(√3 sinθ + 1) - (√3 sinθ + 1)²) * √3 cosθ dθ.
Simplify the expression: (√3 sinθ + 1)² * √(3 + 2(√3 sinθ + 1) - (√3 sinθ + 1)²) * √3 cosθ = (√3 sinθ + 1)² * √(3 + 2√3 sinθ + 2 - 3 sin²θ - 2√3 sinθ - 1) * √3 cosθ.
After simplifying further, integrate with respect to θ to obtain (1/6) [(√3 sinθ + 1)³ + C]. This expression represents the antiderivative of the original function. Hence, the final result of the integral is (1/6) [(√3 sinθ + 1)³ + C], where C is the constant of integration.