## Understanding the Integral of (a^2 - x^2)^1/2

The integral of (a^2 - x^2)^1/2 is a common integral that appears in various fields, including physics, engineering, and mathematics. This integral represents the area under the curve of the function (a^2 - x^2)^1/2, and it has a closed-form solution involving trigonometric functions.

### The Formula

The integral of (a^2 - x^2)^1/2 is given by:

**∫(a^2 - x^2)^1/2 dx = (1/2) * [x * (a^2 - x^2)^1/2 + a^2 * arcsin(x/a)] + C**

where:

**a**is a constant**x**is the variable of integration**C**is the constant of integration

### Derivation of the Formula

The derivation of this formula involves using trigonometric substitution and the following steps:

**Substitution:**Let x = a * sin(θ), then dx = a * cos(θ) dθ**Simplify:**Substitute x and dx in the integral, and simplify using trigonometric identities.**Integration:**Integrate the simplified expression with respect to θ.**Substitute back:**Substitute θ back in terms of x using the original substitution.

### Applications

The integral of (a^2 - x^2)^1/2 has numerous applications, including:

**Finding the area of a semicircle:**The integral represents the area of a semicircle with radius 'a'.**Calculating the volume of a sphere:**The integral is used in calculating the volume of a sphere.**Solving problems in physics and engineering:**The integral appears in various physics and engineering problems related to motion, oscillations, and waves.

### Example

Let's calculate the definite integral of (a^2 - x^2)^1/2 from 0 to a:

**∫(a^2 - x^2)^1/2 dx = (1/2) * [x * (a^2 - x^2)^1/2 + a^2 * arcsin(x/a)] + C**

Evaluating at the limits of integration:

**[(1/2) * [a * (a^2 - a^2)^1/2 + a^2 * arcsin(a/a)]] - [(1/2) * [0 * (a^2 - 0^2)^1/2 + a^2 * arcsin(0/a)]]**

Simplifying:

**[(1/2) * (a^2 * π/2)] - 0 = (1/4) * π * a^2**

This result represents the area of a quarter circle with radius 'a'.

### Conclusion

The integral of (a^2 - x^2)^1/2 is a fundamental integral with various applications in different fields. Understanding its derivation and applications allows for solving complex problems involving areas, volumes, and physical phenomena.