Some trigonometric substitutions | Part-7

Integration of \( \sqrt{\frac{a - x}{a + x}} \)

The expression \( \sqrt{\frac{a - x}{a + x}} \) often arises in calculus problems involving arc lengths, trigonometric integrals, or inverse trigonometric forms. Recognizing and transforming this kind of integral can simplify evaluation significantly.

To evaluate:

\[ \int \sqrt{\frac{a - x}{a + x}} \, dx \] We use the substitution: \[ x = a \sin\theta \quad \Rightarrow \quad dx = a \cos\theta \, d\theta \] Then: \[ a - x = a(1 - \sin\theta), \quad a + x = a(1 + \sin\theta) \] So: \[ \sqrt{\frac{a - x}{a + x}} = \sqrt{\frac{1 - \sin\theta}{1 + \sin\theta}} = \tan\left(\frac{\pi}{4} - \frac{\theta}{2}\right) \]

Now substitute into the integral: \[ \int \sqrt{\frac{a - x}{a + x}} \, dx = \int \tan\left(\frac{\pi}{4} - \frac{\theta}{2}\right) \cdot a \cos\theta \, d\theta \] While this can be solved using tangent half-angle identities, a more direct and widely known result (through standard integral tables or reduction techniques) is:

Result:

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

When to Use This:

This integral appears in scenarios such as:

• Calculating arc lengths where the integrand has square roots of rational expressions.
• Evaluating definite integrals with bounds involving trigonometric or inverse trigonometric behavior.
• Simplifying expressions that look like transformations of circle or ellipse equations.