To evaluate ∫ dx / √(x² + 16), which substitution should you use, and why?
Ax = 4 sin(θ), because the expression under the radical has the form a² − x²
Bx = 4 tan(θ), because the expression under the radical has the form a² + x²
Cx = 4 sec(θ), because the expression under the radical has the form x² − a²
Dx = 4 cos(θ), because cosine simplifies square roots of sums
The radical √(x² + 16) has the form √(x² + a²) with a = 4. This matches the second case: substitute x = a tan(θ), so that x² + a² = a²(1 + tan²θ) = a²sec²θ, and the square root becomes a|sec θ|. Using sin(θ) is the most common error — that substitution is for √(a² − x²), where the x² is subtracted. Each radical form maps to exactly one Pythagorean identity.
Question 2 Multiple Choice
After completing a trigonometric substitution and integrating in θ, a student leaves the answer in terms of θ. What error has the student made?
ANo error — θ is a valid variable since the substitution replaced x with a trig function of θ
BThe student should have differentiated, not integrated, after substituting
CThe student forgot to convert the answer back to x using a reference triangle
DThe student should have used u-substitution instead, which doesn't require back-conversion
The original integral is a function of x, so the final answer must also be expressed in x. After integrating in θ, you must use a reference triangle to convert every trig function of θ back to an algebraic expression in x. For example, if x = a tan θ, then the reference triangle has opposite side x, adjacent side a, and hypotenuse √(x² + a²) — from which sin θ = x/√(x² + a²), cos θ = a/√(x² + a²), etc. Forgetting this step is one of the most common errors in applying trig substitution.
Question 3 True / False
The substitution x = a sin(θ) works on the radical √(a² − x²) because it converts the expression under the radical into a²cos²θ via the Pythagorean identity sin²θ + cos²θ = 1.
TTrue
FFalse
Answer: True
Substituting x = a sin θ gives a² − x² = a² − a²sin²θ = a²(1 − sin²θ) = a²cos²θ. So √(a² − x²) = a|cos θ|, eliminating the square root entirely. This is exactly why trigonometric substitution works: each radical form corresponds to one Pythagorean identity that collapses the square root into a trig function. The entire technique is built on this identity-matching.
Question 4 True / False
If the expression under the radical is not already in the form a² ± x² or x² − a², trigonometric substitution cannot be applied.
TTrue
FFalse
Answer: False
When the quadratic under the radical is not in standard form, you complete the square first to put it into one of the three standard forms. For example, √(2x − x²) = √(1 − (x−1)²) after completing the square — now it fits the form √(a² − u²) with a = 1 and u = x − 1, and you substitute u = sin θ. Completing the square is a preparatory step, not a separate technique; trig substitution then proceeds normally.
Question 5 Short Answer
Why does trigonometric substitution succeed at eliminating square roots of quadratic expressions, when ordinary algebraic methods and u-substitution fail?
Think about your answer, then reveal below.
Model answer: Trigonometric substitution works by exploiting the Pythagorean identities (sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, sec²θ − 1 = tan²θ) to collapse the square root. By substituting x = a·trig(θ), the expression under the radical becomes a perfect square of a trig function, so the square root disappears. U-substitution requires the integrand to contain a function and its derivative simultaneously — a condition that square roots of quadratics don't satisfy. Only the Pythagorean identities provide the algebraic relationship needed to eliminate these radicals.
The key insight is that trig substitution is not guesswork — it is a deliberate exploitation of known identities. The three cases (sin, tan, sec) correspond exactly to the three Pythagorean identities, each designed to eliminate one of the three radical forms. No other substitution has this property. Once the radical is eliminated, the resulting trig integral — which may still require techniques from trig integrals — can be computed. The back-conversion via reference triangle then restores the answer to the original variable x.