Choose a branch before simplifying a radical

Try

The expression √(9−x²) describes the upper semicircle of radius 3. If x=3sin θ, the quantity under the radical becomes 9cos²θ. Is its square root automatically 3cos θ? Only when cosine is nonnegative. A substitution must specify a branch before removing an absolute value.

Predict

Predict which identity matches each radical: 1−sin²θ=cos²θ, 1+tan²θ=sec²θ, or sec²θ−1=tan²θ. Match the structure first, then decide a range for θ that represents the original x interval without ambiguity.

Experiment

For positive a, use this planning table. It describes common choices, with branch restrictions still to be checked in each problem.

Radical form	Substitution	Result before branch simplification
√(a²−x²)	x=a sin θ	a·∣cos θ∣
√(a²+x²)	x=a tan θ	a·∣sec θ∣
√(x²−a²)	x=a sec θ	a·∣tan θ∣

For ∫₀³√(9−x²)dx, choose θ from 0 to π/2. Then the radical is 3cos θ and dx=3cos θ dθ, making the integral 9∫₀^(π/2)cos²θ dθ=9π/4. Draw the quarter disk to verify the result geometrically: one quarter of π·3² is the same value.

Understand

The square root function returns a nonnegative number. In general √(z²)=|z|, not z. For a sine substitution, choosing θ in [−π/2,π/2] makes cosine nonnegative and makes sine one-to-one. For a tangent substitution, the open interval (−π/2,π/2) covers all real x and has positive secant. A secant substitution needs a branch matched to either x≥a or x≤−a; one branch cannot be silently used for both.

The differential matters as much as the radical. With x=2tan θ, dx=2sec²θ dθ and x²+4=4sec²θ. Therefore ∫dx/(x²+4) becomes (1/2)∫dθ, yielding (1/2)arctan(x/2)+C. Forgetting the scale factor in dx would produce a primitive whose derivative is wrong by a constant factor.

For x>2 in ∫dx/√(x²−4), use x=2sec θ with 0<θ<π/2. Both secant and tangent are positive. The radical becomes 2tan θ and the differential 2sec θ tan θ dθ, leaving ∫sec θ dθ. A primitive is ln|sec θ+tan θ|. Returning to x gives ln(x+√(x²−4))+C after absorbing the constant −ln 2. Verify it by differentiation on x>2; extending to other branches requires a separate sign and domain analysis.

Returning to x can use a reference triangle, but its side lengths must be nonnegative and consistent with the chosen branch. The triangle is a representation of a ratio identity, not permission to ignore a negative x in another domain. In definite integrals, transformed angle bounds can avoid back-substitution entirely if all factors and limits use θ consistently.

Some radicals first require completing the square. Others have simpler algebraic or hyperbolic substitutions. Trigonometric substitution is one justified method, not an instruction to replace every quadratic with trigonometric expressions even when ordinary substitution suffices.

Apply

For each practice task, write the chosen angle interval, transformed differential and simplified radical. Explain every removed absolute value. Compare the quarter-circle integral with its geometric area, and differentiate the logarithmic result on its stated domain.

Continue

Next consider intervals with infinite endpoints or singularities. An antiderivative is useful, but a limit—not an ordinary endpoint substitution—determines whether the accumulation is finite.