Question

Evaluate: \(\int \sqrt{3 - 2x - x^2} dx\).

Hint:

When completing the square for a quadratic \(ax^2+bx+c\), it's often easiest to factor out 'a' first, especially if it's negative. Here, we factored out -1 from \(-x^2-2x\) to get \( -(x^2+2x) \), making it easier to see that we need to add and subtract 1 inside the parenthesis.

Answer 1

Step 1: Understanding the Concept:
This is an integral of a square root of a quadratic expression. The standard technique is to complete the square for the quadratic expression inside the root to transform it into the form \(\sqrt{a^2 - u^2}\), \(\sqrt{u^2 - a^2}\), or \(\sqrt{u^2 + a^2}\). This allows the use of a standard integration formula.
Step 2: Key Formula or Approach:
1. Complete the square for the quadratic \(3 - 2x - x^2\).
2. Rewrite the integral in the standard form \(\int \sqrt{a^2 - u^2} du\).
3. Apply the standard integration formula: \[ \int \sqrt{a^2 - u^2} du = \frac{u}{2}\sqrt{a^2 - u^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{u}{a}\right) + C \] Step 3: Detailed Explanation:
First, complete the square for the expression inside the square root: \[ 3 - 2x - x^2 = -(x^2 + 2x - 3) \] \[ = -( (x^2 + 2x + 1) - 1 - 3 ) \] \[ = -( (x+1)^2 - 4 ) \] \[ = 4 - (x+1)^2 \] So, the integral becomes: \[ \int \sqrt{4 - (x+1)^2} dx \] This is now in the form \(\int \sqrt{a^2 - u^2} du\), where: \(a^2 = 4 \implies a = 2\)
\(u = x+1 \implies du = dx\)
Now, apply the standard formula: \[ \int \sqrt{a^2 - u^2} du = \frac{u}{2}\sqrt{a^2 - u^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{u}{a}\right) + C \] Substitute back \(a=2\) and \(u=x+1\): \[ \int \sqrt{4 - (x+1)^2} dx = \frac{x+1}{2}\sqrt{4 - (x+1)^2} + \frac{4}{2}\sin^{-1}\left(\frac{x+1}{2}\right) + C \] \[ = \frac{x+1}{2}\sqrt{3 - 2x - x^2} + 2\sin^{-1}\left(\frac{x+1}{2}\right) + C \] Step 4: Final Answer:
The value of the integral is \(\frac{x+1}{2}\sqrt{3 - 2x - x^2} + 2\sin^{-1}\left(\frac{x+1}{2}\right) + C\).