In this lesson you’ll learn trigonometric substitution, a technique for handling integrals with square roots of quadratic expressions. It turns ugly radicals into clean trig functions.
When you see a square root like a 2 − x 2 \sqrt{a^2 - x^2} a 2 − x 2 x 2 + a 2 \sqrt{x^2 + a^2} x 2 + a 2 x 2 − a 2 \sqrt{x^2 - a^2} x 2 − a 2 x x x
These are the identities that make trig substitution work. You’ll use them constantly in this lesson.
Pythagorean identities
sin 2 θ + cos 2 θ = 1 \sin^2\theta + \cos^2\theta = 1 sin 2 θ + cos 2 θ = 1 1 + tan 2 θ = sec 2 θ 1 + \tan^2\theta = \sec^2\theta 1 + tan 2 θ = sec 2 θ sec 2 θ − 1 = tan 2 θ \sec^2\theta - 1 = \tan^2\theta sec 2 θ − 1 = tan 2 θ Double angle and half-angle (for simplifying after substitution)
sin ( 2 θ ) = 2 sin θ cos θ \sin(2\theta) = 2\sin\theta\cos\theta sin ( 2 θ ) = 2 sin θ cos θ cos 2 θ = 1 + cos 2 θ 2 \cos^2\theta = \frac{1 + \cos 2\theta}{2} cos 2 θ = 2 1 + cos 2 θ sin 2 θ = 1 − cos 2 θ 2 \sin^2\theta = \frac{1 - \cos 2\theta}{2} sin 2 θ = 2 1 − cos 2 θ Key derivatives (you’ll need these for computing dx)
d d θ [ sin θ ] = cos θ d d θ [ tan θ ] = sec 2 θ d d θ [ sec θ ] = sec θ tan θ \frac{d}{d\theta}[\sin\theta] = \cos\theta \qquad \frac{d}{d\theta}[\tan\theta] = \sec^2\theta \qquad \frac{d}{d\theta}[\sec\theta] = \sec\theta\tan\theta d θ d [ sin θ ] = cos θ d θ d [ tan θ ] = sec 2 θ d θ d [ sec θ ] = sec θ tan θ And one integral that shows up in the tan and sec cases
∫ sec θ d θ = ln ∣ sec θ + tan θ ∣ + C \int \sec\theta\,d\theta = \ln|\sec\theta + \tan\theta| + C ∫ sec θ d θ = ln ∣ sec θ + tan θ ∣ + C Each form of square root has a specific substitution.
For a 2 − x 2 \sqrt{a^2 - x^2} a 2 − x 2 x = a sin θ x = a\sin\theta x = a sin θ
a 2 − x 2 = a 2 − a 2 sin 2 θ = a cos θ \sqrt{a^2 - x^2} = \sqrt{a^2 - a^2\sin^2\theta} = a\cos\theta a 2 − x 2 = a 2 − a 2 sin 2 θ = a cos θ For x 2 + a 2 \sqrt{x^2 + a^2} x 2 + a 2 x = a tan θ x = a\tan\theta x = a tan θ
x 2 + a 2 = a 2 tan 2 θ + a 2 = a sec θ \sqrt{x^2 + a^2} = \sqrt{a^2\tan^2\theta + a^2} = a\sec\theta x 2 + a 2 = a 2 tan 2 θ + a 2 = a sec θ For x 2 − a 2 \sqrt{x^2 - a^2} x 2 − a 2 x = a sec θ x = a\sec\theta x = a sec θ
x 2 − a 2 = a 2 sec 2 θ − a 2 = a tan θ \sqrt{x^2 - a^2} = \sqrt{a^2\sec^2\theta - a^2} = a\tan\theta x 2 − a 2 = a 2 sec 2 θ − a 2 = a tan θ
Identify which form is under the radical
Make the substitution (x = a sin θ x = a\sin\theta x = a sin θ a tan θ a\tan\theta a tan θ a sec θ a\sec\theta a sec θ
Compute d x dx d x
Replace everything in the integral (no x x x
Simplify and integrate the trig expression
Draw a reference triangle to convert back to x x x
This is the key step students often skip. After integrating in terms of θ \theta θ x x x
For x = a sin θ x = a\sin\theta x = a sin θ = x = x = x = a = a = a = a 2 − x 2 = \sqrt{a^2 - x^2} = a 2 − x 2
For x = a tan θ x = a\tan\theta x = a tan θ = x = x = x = a = a = a = x 2 + a 2 = \sqrt{x^2 + a^2} = x 2 + a 2
For x = a sec θ x = a\sec\theta x = a sec θ = x = x = x = a = a = a = x 2 − a 2 = \sqrt{x^2 - a^2} = x 2 − a 2
Read off whatever trig functions you need from the triangle.
Case 1: x = a sin θ θ x a √(a² − x²) Case 2: x = a tan θ θ x √(x² + a²) a Case 3: x = a sec θ θ √(x² − a²) x a
Example 1: a 2 − x 2 \sqrt{a^2 - x^2} a 2 − x 2
Evaluate
∫ 9 − x 2 d x \int \sqrt{9 - x^2}\,dx ∫ 9 − x 2 d x Here a = 3 a = 3 a = 3 x = 3 sin θ x = 3\sin\theta x = 3 sin θ d x = 3 cos θ d θ dx = 3\cos\theta\,d\theta d x = 3 cos θ d θ
9 − x 2 = 9 − 9 sin 2 θ = 3 cos θ \sqrt{9 - x^2} = \sqrt{9 - 9\sin^2\theta} = 3\cos\theta 9 − x 2 = 9 − 9 sin 2 θ = 3 cos θ The integral becomes
∫ 3 cos θ ⋅ 3 cos θ d θ = 9 ∫ cos 2 θ d θ \int 3\cos\theta \cdot 3\cos\theta\,d\theta = 9\int \cos^2\theta\,d\theta ∫ 3 cos θ ⋅ 3 cos θ d θ = 9 ∫ cos 2 θ d θ Use the half-angle identity cos 2 θ = ( 1 + cos 2 θ ) / 2 \cos^2\theta = (1 + \cos 2\theta)/2 cos 2 θ = ( 1 + cos 2 θ ) /2
9 ∫ 1 + cos 2 θ 2 d θ = 9 2 ( θ + sin 2 θ 2 ) + C 9\int \frac{1 + \cos 2\theta}{2}\,d\theta = \frac{9}{2}\left(\theta + \frac{\sin 2\theta}{2}\right) + C 9 ∫ 2 1 + cos 2 θ d θ = 2 9 ( θ + 2 sin 2 θ ) + C Since sin ( 2 θ ) = 2 sin θ cos θ \sin(2\theta) = 2\sin\theta\cos\theta sin ( 2 θ ) = 2 sin θ cos θ
= 9 2 θ + 9 2 sin θ cos θ + C = \frac{9}{2}\theta + \frac{9}{2}\sin\theta\cos\theta + C = 2 9 θ + 2 9 sin θ cos θ + C Now convert back. From x = 3 sin θ x = 3\sin\theta x = 3 sin θ sin θ = x / 3 \sin\theta = x/3 sin θ = x /3 θ = arcsin ( x / 3 ) \theta = \arcsin(x/3) θ = arcsin ( x /3 ) cos θ = 9 − x 2 / 3 \cos\theta = \sqrt{9 - x^2}/3 cos θ = 9 − x 2 /3
= 9 2 arcsin ( x 3 ) + x 9 − x 2 2 + C = \frac{9}{2}\arcsin\!\left(\frac{x}{3}\right) + \frac{x\sqrt{9 - x^2}}{2} + C = 2 9 arcsin ( 3 x ) + 2 x 9 − x 2 + C Example 2: x 2 + a 2 \sqrt{x^2 + a^2} x 2 + a 2
Evaluate
∫ d x x 2 + 4 \int \frac{dx}{\sqrt{x^2 + 4}} ∫ x 2 + 4 d x Here a = 2 a = 2 a = 2 x = 2 tan θ x = 2\tan\theta x = 2 tan θ d x = 2 sec 2 θ d θ dx = 2\sec^2\theta\,d\theta d x = 2 sec 2 θ d θ
x 2 + 4 = 4 tan 2 θ + 4 = 2 sec θ \sqrt{x^2 + 4} = \sqrt{4\tan^2\theta + 4} = 2\sec\theta x 2 + 4 = 4 tan 2 θ + 4 = 2 sec θ The integral becomes
∫ 2 sec 2 θ d θ 2 sec θ = ∫ sec θ d θ = ln ∣ sec θ + tan θ ∣ + C \int \frac{2\sec^2\theta\,d\theta}{2\sec\theta} = \int \sec\theta\,d\theta = \ln|\sec\theta + \tan\theta| + C ∫ 2 sec θ 2 sec 2 θ d θ = ∫ sec θ d θ = ln ∣ sec θ + tan θ ∣ + C From the reference triangle with x = 2 tan θ x = 2\tan\theta x = 2 tan θ tan θ = x / 2 \tan\theta = x/2 tan θ = x /2 x 2 + 4 \sqrt{x^2 + 4} x 2 + 4 sec θ = x 2 + 4 / 2 \sec\theta = \sqrt{x^2 + 4}/2 sec θ = x 2 + 4 /2
= ln ∣ x 2 + 4 2 + x 2 ∣ + C = ln ∣ x + x 2 + 4 ∣ + C 1 = \ln\left|\frac{\sqrt{x^2+4}}{2} + \frac{x}{2}\right| + C = \ln\left|x + \sqrt{x^2+4}\right| + C_1 = ln 2 x 2 + 4 + 2 x + C = ln x + x 2 + 4 + C 1 (The ln(2) got absorbed into the constant.)
Example 3: x 2 − a 2 \sqrt{x^2 - a^2} x 2 − a 2
Evaluate
∫ d x x 2 − 9 ( x > 3 ) \int \frac{dx}{\sqrt{x^2 - 9}} \qquad (x > 3) ∫ x 2 − 9 d x ( x > 3 ) Here a = 3 a = 3 a = 3 x = 3 sec θ x = 3\sec\theta x = 3 sec θ d x = 3 sec θ tan θ d θ dx = 3\sec\theta\tan\theta\,d\theta d x = 3 sec θ tan θ d θ
x 2 − 9 = 9 sec 2 θ − 9 = 3 tan θ \sqrt{x^2 - 9} = \sqrt{9\sec^2\theta - 9} = 3\tan\theta x 2 − 9 = 9 sec 2 θ − 9 = 3 tan θ The integral becomes
∫ 3 sec θ tan θ d θ 3 tan θ = ∫ sec θ d θ = ln ∣ sec θ + tan θ ∣ + C \int \frac{3\sec\theta\tan\theta\,d\theta}{3\tan\theta} = \int \sec\theta\,d\theta = \ln|\sec\theta + \tan\theta| + C ∫ 3 tan θ 3 sec θ tan θ d θ = ∫ sec θ d θ = ln ∣ sec θ + tan θ ∣ + C From the reference triangle: sec θ = x / 3 \sec\theta = x/3 sec θ = x /3 tan θ = x 2 − 9 / 3 \tan\theta = \sqrt{x^2 - 9}/3 tan θ = x 2 − 9 /3
= ln ∣ x 3 + x 2 − 9 3 ∣ + C = ln ∣ x + x 2 − 9 ∣ + C 1 = \ln\left|\frac{x}{3} + \frac{\sqrt{x^2-9}}{3}\right| + C = \ln\left|x + \sqrt{x^2-9}\right| + C_1 = ln 3 x + 3 x 2 − 9 + C = ln x + x 2 − 9 + C 1 Trig substitution shows up in problems involving circles, ellipses, and other curved geometry.
Computing the area of an ellipse or a circular segment requires integrals with a 2 − x 2 \sqrt{a^2 - x^2} a 2 − x 2
In physics, gravitational and electric field calculations for continuous charge distributions often produce these radical forms
Arc length and surface area formulas frequently lead to 1 + ( d y / d x ) 2 \sqrt{1 + (dy/dx)^2} 1 + ( d y / d x ) 2 x 2 + a 2 \sqrt{x^2 + a^2} x 2 + a 2
In engineering, stress analysis on curved beams and fluid flow around cylindrical objects use these integrals
Trig substitution is really just pattern matching. Look at the radical, pick the substitution from the three cases, and follow the steps. The reference triangle is your best friend for converting back. If you can draw the triangle correctly, the back-substitution is just reading off sides. Don’t let the θ \theta θ
For an integral containing $\sqrt{a^2 - x^2}$, the correct substitution is A. $x = a\tan\theta$ B. $x = a\sec\theta$ C. $x = a\sin\theta$ D. $x = a\cos\theta$
For an integral containing $\sqrt{x^2 + a^2}$, the correct substitution is A. $x = a\sin\theta$ B. $x = a\tan\theta$ C. $x = a\sec\theta$ D. $x = a\cos\theta$
After integrating with trig substitution, you convert back to $x$ using A. L'Hopital's rule B. A reference right triangle C. Another round of substitution D. The quadratic formula
$\int \sqrt{9 - x^2}\,dx$ evaluates to A. $\frac{9}{2}\arcsin\!\left(\frac{x}{3}\right) - \frac{x\sqrt{9-x^2}}{2} + C$ B. $\frac{9}{2}\arctan\!\left(\frac{x}{3}\right) + \frac{x\sqrt{9-x^2}}{2} + C$ C. $\frac{9}{2}\arcsin\!\left(\frac{x}{3}\right) + \frac{x\sqrt{9-x^2}}{2} + C$ D. $3\arcsin\!\left(\frac{x}{3}\right) + x\sqrt{9-x^2} + C$
After substituting $x = 3\sec\theta$ into $\int \frac{dx}{\sqrt{x^2 - 9}}$, the integral simplifies to A. $\int \cos\theta\,d\theta$ B. $\int \sec\theta\,d\theta$ C. $\int \tan\theta\,d\theta$ D. $\int \csc\theta\,d\theta$
Retry Quiz
