### Understanding the Integral The integral we're dealing with is: \[ \int \frac{\sqrt{a^2 - x^2}}{\sqrt{a^2 + x^2}} \, dx \] At first glance, it's a ratio of two square roots. The numerator is \(\sqrt{a^2 - x^2}\), which resembles the form found in integrals involving trigonometric substitutions, especially those related to the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \). The denominator is \(\sqrt{a^2 + x^2}\), which might suggest a hyperbolic substitution or another trigonometric identity. ### Initial Approach: Trigonometric Substitution Given the presence of \( \sqrt{a^2 - x^2} \), a common substitution is \( x = a \sin \theta \). Let's try that: Let \( x = a \sin \theta \), then \( dx = a \cos \theta \, d\theta \). Substituting into the integral: \[ \sqrt{a^2 - x^2} = \sqrt{a^2 - a^2 \sin^2 \theta} = a \cos \theta \] \[ \sqrt{a^2 + x^2} = \sqrt{a^2 + a^2 \sin^2 \theta} = a \sqrt{1 + \sin^2 \theta} \] So the integral becomes: \[ \int \frac{a \cos \theta}{a \sqrt{1 + \sin^2 \theta}} \cdot a \cos \theta \, d\theta = a \int \frac{\cos^2 \theta}{\sqrt{1 + \sin^2 \theta}} \, d\theta \] This seems to complicate things because we now have \( \cos^2 \theta \) over \( \sqrt{1 + \sin^2 \theta} \), which isn't straightforward to integrate. ### Alternative Approach: Hyperbolic Substitution Given that the denominator is \( \sqrt{a^2 + x^2} \), perhaps a hyperbolic substitution would help. The hyperbolic identity is \( \cosh^2 \phi - \sinh^2 \phi = 1 \), so \( \sqrt{a^2 + x^2} \) suggests \( x = a \sinh \phi \). Let \( x = a \sinh \phi \), then \( dx = a \cosh \phi \, d\phi \). Substituting: \[ \sqrt{a^2 - x^2} = \sqrt{a^2 - a^2 \sinh^2 \phi} = a \sqrt{1 - \sinh^2 \phi} = a \sqrt{-\cosh^2 \phi + 2} \quad \text{(This seems problematic)} \] Wait, \( \sqrt{1 - \sinh^2 \phi} \) is \( \sqrt{-\cosh^2 \phi + 2} \), which involves imaginary numbers unless \( \sinh \phi \leq 1 \). This might not simplify things nicely. ### Another Idea: Euler Substitution For integrals involving \( \sqrt{a^2 + x^2} \) and \( \sqrt{a^2 - x^2} \), an Euler substitution might be applicable. The general idea is to substitute to eliminate the square roots. Let’s consider the substitution: \[ \sqrt{a^2 + x^2} = t - x \] Then: \[ a^2 + x^2 = t^2 - 2 t x + x^2 \implies a^2 = t^2 - 2 t x \implies x = \frac{t^2 - a^2}{2 t} \] Compute \( dx \): \[ dx = \frac{2 t \cdot 2 t - 2 (t^2 - a^2)}{4 t^2} dt = \frac{4 t^2 - 2 t^2 + 2 a^2}{4 t^2} dt = \frac{2 t^2 + 2 a^2}{4 t^2} dt = \frac{t^2 + a^2}{2 t^2} dt \] Now, express \( \sqrt{a^2 - x^2} \): \[ x = \frac{t^2 - a^2}{2 t}, \quad a^2 - x^2 = a^2 - \left( \frac{t^2 - a^2}{2 t} \right)^2 = \frac{4 a^2 t^2 - (t^4 - 2 a^2 t^2 + a^4)}{4 t^2} = \frac{-t^4 + 6 a^2 t^2 - a^4}{4 t^2} \] This seems messy, and taking the square root of this expression doesn't appear to