integrate: square root of (4-x^2)
I think its 2-x^1, but I can't be sure as I learned it in precalculus.
No to matt's answer. It requires integration by substitution and one other trick. Let u = 4 - x^2.
x = sqrt(4-u) dx = -(1/2)/sqrt(4-u)
That makes the integral that of
-(1/2) u^(1/2)/sqrt(4-u)
Perhaps that can be integrated by parts.
The answer (from my table of integrals) is (1/2)[x*sqrt(4-x^2) -4 sin^-1(x/2)]
I just don't have time to do it. perhaps Reiny or Damon will come to the resuce
To integrate √(4 - x^2), we can use a trigonometric substitution. Let's follow these steps:
1. Rewrite the expression as √(2^2 - x^2).
2. Notice that this resembles the standard form of a circle: x^2 + y^2 = r^2. In our case, the radius is 2.
3. To make a substitution, let x = 2sinθ, which means dx = 2cosθdθ.
4. Now, substitute x and dx with the appropriate trigonometric expressions in terms of θ. The square root becomes √(4 - (2sinθ)^2) = √(4 - 4sin^2θ) = √(4cos^2θ) = 2cosθ.
5. Substitute dx as well: dx = 2cosθdθ.
6. Rewrite the integral using the substitution: ∫2cosθ * 2cosθdθ.
7. Simplify: ∫4cos^2θdθ.
8. Apply the identity cos^2θ = (1 + cos2θ)/2.
9. Rewrite the integral: ∫(2 + 2cos2θ)dθ.
10. Integrate each term separately. The integral of 2 with respect to θ is 2θ. Use the trigonometric identity sin2θ = (1 - cos2θ)/2 to integrate 2cos2θ, which becomes sin2θ.
11. Rewrite the integral: 2θ + sin2θ + C, where C is the constant of integration.
Therefore, the indefinite integral of √(4 - x^2) is 2θ + sin2θ + C, where θ is determined by the substitution x = 2sinθ.