Mathwords logoReference LibraryMathwords

Trig Substitution

Trig Substitution

A method for computing integrals often used when the integrand contains expressions of the form a2 – x2, a2 + x2, or x2 – a2.

 

Table showing trig substitutions: a²–x² use x=a sinθ; a²+x² use x=a tanθ; x²–a² use x=a secθ
Right triangle with hypotenuse 2, vertical side x, horizontal side √(4−x²), where x/2 = sinθ, illustrating trig substitution.

 

 

See also

u-substitution

Key Formula

For a2−x2:x=asin⁡θ,dx=acos⁡θ dθFor a2+x2:x=atan⁡θ,dx=asec⁡2θ dθFor x2−a2:x=asec⁡θ,dx=asec⁡θtan⁡θ dθ\text{For } \sqrt{a^2 - x^2}:\quad x = a\sin\theta,\quad dx = a\cos\theta\,d\theta \\ \text{For } \sqrt{a^2 + x^2}:\quad x = a\tan\theta,\quad dx = a\sec^2\theta\,d\theta \\ \text{For } \sqrt{x^2 - a^2}:\quad x = a\sec\theta,\quad dx = a\sec\theta\tan\theta\,d\theta
Where:
  • xx = The original variable of integration
  • aa = A positive constant appearing in the expression under the radical
  • θ\theta = The new variable introduced through the trigonometric substitution

Worked Example

Problem: Evaluate ∫ √(9 − x²) dx.
Step 1: Identify the form. The integrand contains √(a² − x²) with a = 3, so use the substitution x = 3 sin θ.
x=3sin⁡θ,dx=3cos⁡θ dθx = 3\sin\theta, \quad dx = 3\cos\theta\,d\theta
Step 2: Substitute and simplify the radical using the Pythagorean identity sin²θ + cos²θ = 1.
9−x2=9−9sin⁡2θ=3cos⁡θ\sqrt{9 - x^2} = \sqrt{9 - 9\sin^2\theta} = 3\cos\theta
Step 3: Rewrite the entire integral in terms of θ.
∫3cos⁡θ⋅3cos⁡θ dθ=9∫cos⁡2θ dθ\int 3\cos\theta \cdot 3\cos\theta\,d\theta = 9\int \cos^2\theta\,d\theta
Step 4: Apply the half-angle identity cos²θ = (1 + cos 2θ)/2 and integrate.
9∫1+cos⁡2θ2 dθ=92θ+94sin⁡2θ+C9\int \frac{1 + \cos 2\theta}{2}\,d\theta = \frac{9}{2}\theta + \frac{9}{4}\sin 2\theta + C
Step 5: Convert back to x. Since x = 3 sin θ, we have θ = arcsin(x/3). Also, sin 2θ = 2 sin θ cos θ = 2·(x/3)·(√(9 − x²)/3) = 2x√(9 − x²)/9.
92arcsin⁡ ⁣(x3)+x9−x22+C\frac{9}{2}\arcsin\!\left(\frac{x}{3}\right) + \frac{x\sqrt{9 - x^2}}{2} + C
Answer: 92arcsin⁡ ⁣(x3)+x9−x22+C\frac{9}{2}\arcsin\!\left(\frac{x}{3}\right) + \frac{x\sqrt{9 - x^2}}{2} + C

Another Example

This example uses the x = a sec θ substitution (for √(x² − a²)), whereas the first example used x = a sin θ. It also shows how trig substitution handles a rational integrand with a radical in the denominator.

Problem: Evaluate ∫ 1/(x²√(x² − 4)) dx.
Step 1: Identify the form. The integrand contains √(x² − a²) with a = 2, so use x = 2 sec θ.
x=2sec⁡θ,dx=2sec⁡θtan⁡θ dθx = 2\sec\theta, \quad dx = 2\sec\theta\tan\theta\,d\theta
Step 2: Simplify the radical using the identity sec²θ − 1 = tan²θ.
x2−4=4sec⁡2θ−4=2tan⁡θ\sqrt{x^2 - 4} = \sqrt{4\sec^2\theta - 4} = 2\tan\theta
Step 3: Substitute everything into the integral and simplify.
∫2sec⁡θtan⁡θ dθ(4sec⁡2θ)(2tan⁡θ)=∫dθ4sec⁡θ=14∫cos⁡θ dθ\int \frac{2\sec\theta\tan\theta\,d\theta}{(4\sec^2\theta)(2\tan\theta)} = \int \frac{d\theta}{4\sec\theta} = \frac{1}{4}\int \cos\theta\,d\theta
Step 4: Integrate and convert back. Since x = 2 sec θ, we have cos θ = 2/x and sin θ = √(x² − 4)/x.
14sin⁡θ+C=x2−44x+C\frac{1}{4}\sin\theta + C = \frac{\sqrt{x^2 - 4}}{4x} + C
Answer: x2−44x+C\frac{\sqrt{x^2 - 4}}{4x} + C

Frequently Asked Questions

How do you know which trig substitution to use?
Match the expression under (or inside) the radical to one of three patterns: use x = a sin θ for √(a² − x²), use x = a tan θ for √(a² + x²), and use x = a sec θ for √(x² − a²). The key is recognizing which Pythagorean identity will eliminate the square root. Even if no radical is visible, completing the square may reveal one of these forms.
What is the difference between trig substitution and u-substitution?
In u-substitution, you set u equal to some function of x already present in the integrand, aiming to match du with the remaining factors. In trig substitution, you replace x itself with a trigonometric expression (like a sin θ) to exploit a Pythagorean identity. Trig substitution is specifically designed for integrands involving a² ± x² or x² − a² under a radical, whereas u-substitution is a more general technique.
Do you always need a square root to use trig substitution?
No. Trig substitution also works when the integrand contains expressions like 1/(a² + x²)² or (a² − x²)^(3/2) without an explicit square root sign. Any time you see a sum or difference of squares that resists simpler methods, trig substitution may help. Completing the square can also transform an integrand into one of the standard forms.

Trig Substitution vs. u-Substitution

Trig Substitutionu-Substitution
Core ideaReplace x with a trig function of θ to exploit a Pythagorean identityReplace a sub-expression with u to simplify the chain rule in reverse
Typical triggerIntegrand contains √(a² − x²), √(a² + x²), or √(x² − a²)Integrand contains a function and its derivative as a factor
Substitution directionx is expressed in terms of the new variable θu is expressed in terms of the old variable x
Back-substitutionUse a right triangle or inverse trig to return to xDirectly replace u with its expression in x
ComplexityOften requires additional identities (half-angle, double-angle)Usually a single direct substitution

Why It Matters

Trig substitution appears frequently in Calculus II courses and is essential on AP Calculus BC exams. It is the standard method for computing arc length, surface area, and many physics integrals (such as gravitational or electric field calculations) that naturally produce expressions with a² ± x². Mastering it also deepens your understanding of how Pythagorean identities connect algebra and trigonometry.

Common Mistakes

Mistake: Forgetting to change dx when substituting.
Correction: When you replace x with a trig expression, you must also differentiate to express dx in terms of dθ. For example, if x = a sin θ, then dx = a cos θ dθ — leaving dx unchanged will produce an incorrect integral.
Mistake: Returning the answer in terms of θ instead of converting back to x.
Correction: The original integral is in terms of x, so your final answer must be too. Draw a right triangle labeled with your substitution (e.g., opposite = x, hypotenuse = a for x = a sin θ) and use it to express every trig function of θ as an algebraic expression in x.

Related Terms