Some trigonometric substitutions | Part-4

Integration of \( \sqrt{a^2 - x^2} \) using Trigonometric Substitution

The integral \( \int \sqrt{a^2 - x^2} \, dx \) commonly arises in geometry, physics, and calculus—particularly in problems involving arc length, area under curves, and circle-related expressions. It fits a classic form suitable for trigonometric substitution.

When the integrand involves \( \sqrt{a^2 - x^2} \), we use the substitution: \( x = a \sin\theta \quad\) or \(\quad x = a \cos\theta \) Here, we choose \( x = a \sin\theta \), which leads to: \( dx = a \cos\theta \, d\theta \quad\) and \(\quad \sqrt{a^2 - x^2} = \sqrt{a^2 - a^2 \sin^2\theta} = a \cos\theta \)

Substituting into the integral: \[ \int \sqrt{a^2 - x^2} \, dx = \int a \cos\theta \cdot a \cos\theta \, d\theta = a^2 \int \cos^2\theta \, d\theta \] Apply the identity: \[ \cos^2\theta = \frac{1 + \cos(2\theta)}{2} \] So the integral becomes: \[ \frac{a^2}{2} \int (1 + \cos(2\theta)) \, d\theta = \frac{a^2}{2} \left( \int 1\, d\theta + \int \cos(2\theta)\, d\theta \right) = \frac{a^2}{2} \left( \theta + \frac{1}{2} \sin(2\theta) \right) + C \]

Now revert to \( x \) using \( x = a \sin\theta \Rightarrow \theta = \sin^{-1}\left( \frac{x}{a} \right) \). Also: \[ cos\theta = \frac{\sqrt{a^{2} - x^{2}}}{a}, \quad \] and using \[\sin(2\theta) = 2 \sin\theta \cos\theta,\] we get \[\sin(2\theta) = \frac{2x \sqrt{a^{2} - x^{2}}}{a^{2}} \] Plugging this back in: \[ \frac{a^{2}}{2} \left( \theta + \frac{\sin(2\theta)}{2} \right) = \frac{a^{2}}{2} \sin^{-1}\left( \frac{x}{a} \right) + \frac{x \sqrt{a^{2} - x^{2}}}{2} \]

Final Result:

\[ \int \sqrt{a^{2} - x^{2}} \, dx = \frac{x \sqrt{a^{2} - x^{2}}}{2} + \frac{a^{2} \sin^{-1}\left( \frac{x}{a} \right)}{2} + C \]