Question

Evaluate the integral \( \int \frac{1}{\sqrt{4x - x^2}} \, dx \).

Hint:

For integrals involving square roots of quadratic expressions, use trigonometric substitution to simplify the integrand. Completing the square often helps in setting up the substitution.

Answer 1

Step 1: Rewrite the quadratic expression inside the square root. We can complete the square for \( 4x - x^2 \): \[ 4x - x^2 = 4 - (x - 2)^2. \] Thus, the integral becomes: \[ \int \frac{1}{\sqrt{4 - (x - 2)^2}} \, dx. \] Step 2: Use the standard trigonometric substitution \( x - 2 = 2 \sin \theta \). Then, \( dx = 2 \cos \theta \, d\theta \), and the expression inside the square root becomes: \[ 4 - (x - 2)^2 = 4 - 4 \sin^2 \theta = 4 \cos^2 \theta. \] So, the integral simplifies to: \[ \int \frac{2 \cos \theta}{2 \cos \theta} \, d\theta = \int d\theta. \] Step 3: Integrating \( d\theta \), we get: \[ \theta + C. \] Step 4: Substitute \( \theta = \sin^{-1} \left( \frac{x - 2}{2} \right) \), yielding: \[ \sin^{-1} \left( \frac{x - 2}{2} \right) + C. \]