Trigonometric Substitution

A radical of the form \sqrt{a^2 - x^2} signals the trigonometric substitution x = a\sin\theta, the canonical move that trades an awkward algebraic root for a clean trigonometric identity. The substitution works because the Pythagorean identity 1 - \sin^2\theta = \cos^2\theta lets the radical simplify to a single cosine. Carrying it out, dx = a\cos\theta,d\theta and \sqrt{a^2 - x^2} = a\cos\theta, so the integral becomes \int \frac{a\cos\theta,d\theta}{a\cos\theta} = \int d\theta = \theta + C. Since \theta = \arcsin(x/a), the antiderivative is \arcsin(x/a) + C. Option \arccos(x/a) differs from the answer only by an additive constant but is conventionally not the standard sign-aligned form. Option with \arctan corresponds to the denominator a^2 + x^2, a different radical-free quadratic requiring a tangent substitution. Option with the logarithm matches the radical \sqrt{x^2 - a^2}, which calls for a secant or hyperbolic substitution instead. This sine substitution is a staple of the JEE Advanced toolkit. As a final plausibility check, differentiating \arcsin(x/a) gives \frac{1}{a}\cdot\frac{1}{\sqrt{1 - (x/a)^2}} = \frac{1}{\sqrt{a^2 - x^2}}, which is exactly the integrand and valid precisely on the domain |x| < a where the radical is real.