Integrating √(4x - x^2) with Trigonometric Substitution

The integral ∫√(4x - x^2)dx can be solved using trigonometric substitution. This method is particularly useful when dealing with integrands that involve expressions of the form √(a^2 - u^2).

Problem Statement

Consider the integral (I int sqrt{4x - x^2}dx). Our goal is to evaluate this integral using a trigonometric substitution technique.

Step 1: Completing the Square

The first step is to complete the square for the expression under the square root:

(4x - x^2 4 - (x^2 4x))

4 - (x^2 4x 4 - 4) 4 - (x 2)^2)

4 - (x - 2)^2)

Therefore, the integral can be rewritten as:

(I int sqrt{4 - (x - 2)^2}dx)

Step 2: Trigonometric Substitution

Next, we use the substitution (x - 2 2sintheta), which gives us:

(dx 2costheta dtheta)

Substituting these into the integral, we get:

(I int sqrt{4 - 4sin^2theta} cdot 2costheta dtheta)

2 int 2cos^2theta dtheta)

4 int cos^2theta dtheta)

Using the identity (cos^2theta frac{1 cos 2theta}{2}), we obtain:

(I 4 int frac{1 cos 2theta}{2} dtheta)

2 int (1 cos 2theta) dtheta)

2theta sin 2theta C)

Since (theta sin^{-1}frac{x - 2}{2}), and (sin 2theta 2sinthetacostheta), we have:

(sin 2theta 2 cdot frac{x - 2}{2} cdot frac{sqrt{4x - (x - 2)^2}}{2})

Substituting back, the final result is:

(I 2sin^{-1}frac{x - 2}{2} frac{x - 2}{2} cdot sqrt{4x - x^2} C)

Conclusion

The integral ∫√(4x - x^2)dx is evaluated as:

(boxed{2sin^{-1}frac{x - 2}{2} cdot frac{x - 2}{2} sqrt{4x - x^2} C})

Related Formulas

For the general case ∫√(a^2 - u^2)du, we have the formula:

(int sqrt{a^2 - u^2}du frac{u}{2} sqrt{a^2 - u^2} frac{a^2}{2} sin^{-1} frac{u}{a} C)

For instance, for the specific case where a 2:

(int sqrt{4 - x^2}dx frac{2}{2} sqrt{4 - x^2} frac{4}{2} sin^{-1} frac{x}{2} C sqrt{4 - x^2} 2sin^{-1} frac{x}{2} C)

Proof of the General Formula

To prove the general formula, we can use integration by parts and trigonometric identities. The detailed proof can be found in the video linked here:

∫√a^2u^2 du Proof