Trig Substitution

The square-root barrier

Integrals like ∫√(a²−x²)dx and ∫dx/√(a²+x²) are stuck. Square roots of quadratic expressions don't yield to u-substitution (no inner-function-with-its-derivative pattern), and integration by parts only makes things messier. The square root has to disappear, and that requires an identity. The Pythagorean identity sin²θ + cos²θ = 1, rearranged in three ways, is the key.

The three substitutions

If you see…	Substitute	Then dx becomes	The square root becomes	Identity used
√(a² − x²)	x = a·sin θ	dx = a·cos θ dθ	a·cos θ	1 − sin²θ = cos²θ
√(a² + x²)	x = a·tan θ	dx = a·sec²θ dθ	a·sec θ	1 + tan²θ = sec²θ
√(x² − a²)	x = a·sec θ	dx = a·sec θ·tan θ dθ	a·tan θ	sec²θ − 1 = tan²θ

In each case the Pythagorean identity reduces the square root to a single trig function. The dx absorbs the trig differential. What remains is an integral in θ, usually a clean polynomial in sin, cos, sec, or tan.

How to apply it

1. Identify which of the three square-root forms appears.
2. Substitute x with the matching trig expression. Compute dx.
3. Simplify the square root using the Pythagorean identity.
4. Rewrite the entire integral in θ.
5. Integrate (often a power of sin/cos/sec/tan).
6. Convert back to x using a right-triangle picture matching the substitution.
7. For definite integrals, change the limits to θ at the substitution step.

Worked example — ∫ √(1 − x²) dx (the area under a circle)

This is the canonical trig-sub. The factor √(1 − x²) is the upper half of the unit circle. We want its antiderivative.

x = sin θ         ⟹  dx = cos θ dθ
√(1 − x²) = √(1 − sin²θ) = √cos²θ = cos θ  (principal range)

∫ √(1 − x²) dx  =  ∫ cos θ · cos θ dθ
                =  ∫ cos²θ dθ
                =  ∫ (1 + cos 2θ)/2 dθ
                =  (1/2)θ + (1/4) sin 2θ + C
                =  (1/2)θ + (1/2) sin θ cos θ + C.

Now convert back. The right triangle for x = sin θ: opposite x, hypotenuse 1, adjacent √(1 − x²). So θ = arcsin x, sin θ = x, cos θ = √(1 − x²).

∫ √(1 − x²) dx  =  (1/2) arcsin(x)  +  (1/2) x √(1 − x²)  +  C.

Sanity check — differentiate: (1/2)·(1/√(1−x²)) + (1/2)(√(1−x²) + x·(−x/√(1−x²))) = (1/2)/√(1−x²) + (1/2)(1−x²)/√(1−x²) − (1/2)x²/√(1−x²) = (1/2)·(1 + 1 − x² − x²)/√(1−x²) = (1 − x²)/√(1−x²) = √(1−x²). Matches.

Evaluating from −1 to 1 gives π/2, the area of the upper unit semicircle. The full circle has area π. Trig substitution is how you derive this from first principles.

Worked example — ∫ dx / √(1 + x²)

The form √(a² + x²) with a = 1. Use x = tan θ.

x = tan θ         ⟹  dx = sec²θ dθ
√(1 + x²) = √(1 + tan²θ) = √sec²θ = sec θ

∫ dx / √(1 + x²)  =  ∫ sec²θ dθ / sec θ
                  =  ∫ sec θ dθ
                  =  ln|sec θ + tan θ| + C.

Triangle for x = tan θ — opposite x, adjacent 1, hypotenuse √(1 + x²). So sec θ = √(1 + x²) and tan θ = x.

∫ dx / √(1 + x²)  =  ln| √(1 + x²) + x | + C  =  arcsinh(x) + C.

The closed form is the inverse hyperbolic sine — the same as the answer from the hyperbolic substitution x = sinh u. Both methods land at the same place.

Worked example — ∫ dx / √(x² − 1)

The form √(x² − a²) with a = 1. Use x = sec θ.

x = sec θ         ⟹  dx = sec θ · tan θ dθ
√(x² − 1) = √(sec²θ − 1) = √tan²θ = tan θ  (principal range, x > 1)

∫ dx / √(x² − 1)  =  ∫ sec θ · tan θ dθ / tan θ
                  =  ∫ sec θ dθ
                  =  ln|sec θ + tan θ| + C
                  =  ln| x + √(x² − 1) | + C  =  arccosh(x) + C.

Trig substitution vs other techniques

	Trig substitution	U-substitution	Integration by parts	Hyperbolic substitution
Best when	√(a²±x²) or √(x²−a²) present	Inner function + its derivative	Product of dissimilar factors	√(a²+x²) or √(x²−a²)
Underlying identity	Pythagorean	Chain rule	Product rule	cosh² − sinh² = 1
Order to try	After u-sub fails on a square root	First — fastest	Second	Alternative to tan/sec sub
Bookkeeping	High — three triangles to remember	Low	Medium	High — fewer textbook examples
Closed forms produced	arcsin, arctan, arcsec, log	Whatever ∫f(u)du gives	uv-style mixed forms	arcsinh, arccosh
Speed	Slow (triangle setup)	Fast	Medium	Medium

When trig substitution is the right tool

• Area-of-circle integrals. ∫√(r² − x²) dx — the upper half of a circle of radius r. x = r·sin θ.
• Arc length integrals. The arc-length formula ∫√(1 + (dy/dx)²) dx often produces √(1 + x²) forms that yield to x = tan θ.
• Surfaces of revolution. Surface-area integrals around hyperbolic curves create √(x² − 1) forms — x = sec θ.
• Physics — electric field of a charged rod. Coulomb's-law integrals along a finite line yield exactly the √(a² + x²) form.
• Definite integrals with circular or hyperbolic limits. Changing limits to θ-values simplifies bounded evaluations.
• Probability — Cauchy and circular distributions. Distributions with density 1/(1+x²) come up; arctan is their integral.

Common mistakes

• Picking the wrong substitution. Memorize the three rules. Wrong sub means the square root doesn't simplify.
• Forgetting dx. The dx contributes a·cos θ dθ, a·sec²θ dθ, or a·sec θ·tan θ dθ. Missing this factor is the most common arithmetic error.
• Sign and absolute-value sloppiness. √cos²θ = |cos θ|, not always cos θ. On the principal range of arcsin (−π/2 ≤ θ ≤ π/2), cos θ ≥ 0, so |cos θ| = cos θ. Stay on the principal range and you avoid absolute values; range-cross and you must handle them.
• Not converting back. The answer in θ isn't the final answer — you need to express it in x via the right triangle.
• Wrong triangle. Each substitution has its own triangle (opposite = ?, hypotenuse = ?, adjacent = ?). Mismatched triangles produce wrong arctrig answers.
• Mixing limits and substitutions on definite integrals. Convert limits to θ at the substitution step. Don't evaluate θ-limits with an x-form, or vice versa.

Connections to other techniques

U-substitution. Trig sub is a u-sub where u is a trig function. The framework is identical; only the recognition pattern differs.

Integration by parts. After trig substitution, the resulting trig integral sometimes still requires parts (e.g. ∫sec³θ dθ uses sec·sec² with parts).

Hyperbolic substitution. For √(a²+x²) and √(x²−a²), x = a·sinh u or a·cosh u works in parallel. Hyperbolic sub stays in elementary exponentials, avoiding inverse trig.

Partial fractions. After trig sub yields a rational trig integrand, the Weierstrass substitution t = tan(θ/2) converts it back to a rational function in t, then partial fractions finish the job.

Complex analysis. Many trig-sub problems have shorter solutions via contour integration once you know complex methods; trig sub remains the elementary route.