Trigonometric Substitution

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 $\sqrt{a^2 - x^2}$, $\sqrt{x^2 + a^2}$, or $\sqrt{x^2 - a^2}$, none of the earlier techniques (u-sub, by parts, trig integrals) will simplify it directly. Trig substitution works by replacing $x$ with a trig expression that eliminates the radical via a Pythagorean identity.

These are the identities that make trig substitution work. You’ll use them constantly in this lesson.

Pythagorean identities

$$\sin^2\theta + \cos^2\theta = 1$$ $$1 + \tan^2\theta = \sec^2\theta$$ $$\sec^2\theta - 1 = \tan^2\theta$$

Double angle and half-angle (for simplifying after substitution)

$$\sin(2\theta) = 2\sin\theta\cos\theta$$ $$\cos^2\theta = \frac{1 + \cos 2\theta}{2}$$ $$\sin^2\theta = \frac{1 - \cos 2\theta}{2}$$

Key derivatives (you’ll need these for computing dx)

$$\frac{d}{d\theta}[\sin\theta] = \cos\theta \qquad \frac{d}{d\theta}[\tan\theta] = \sec^2\theta \qquad \frac{d}{d\theta}[\sec\theta] = \sec\theta\tan\theta$$

And one integral that shows up in the tan and sec cases

$$\int \sec\theta\,d\theta = \ln|\sec\theta + \tan\theta| + C$$

Each form of square root has a specific substitution.

For $\sqrt{a^2 - x^2}$, let $x = a\sin\theta$

$$\sqrt{a^2 - x^2} = \sqrt{a^2 - a^2\sin^2\theta} = a\cos\theta$$

For $\sqrt{x^2 + a^2}$, let $x = a\tan\theta$

$$\sqrt{x^2 + a^2} = \sqrt{a^2\tan^2\theta + a^2} = a\sec\theta$$

For $\sqrt{x^2 - a^2}$, let $x = a\sec\theta$

$$\sqrt{x^2 - a^2} = \sqrt{a^2\sec^2\theta - a^2} = a\tan\theta$$

1. Identify which form is under the radical
2. Make the substitution ($x = a\sin\theta$, $a\tan\theta$, or $a\sec\theta$)
3. Compute $dx$
4. Replace everything in the integral (no $x$ should remain)
5. Simplify and integrate the trig expression
6. Draw a reference triangle to convert back to $x$

This is the key step students often skip. After integrating in terms of $\theta$, you need to get back to $x$. Draw a right triangle where

For $x = a\sin\theta$: opposite $= x$, hypotenuse $= a$, adjacent $= \sqrt{a^2 - x^2}$

For $x = a\tan\theta$: opposite $= x$, adjacent $= a$, hypotenuse $= \sqrt{x^2 + a^2}$

For $x = a\sec\theta$: hypotenuse $= x$, adjacent $= a$, opposite $= \sqrt{x^2 - a^2}$

Read off whatever trig functions you need from the triangle.

opposite hypotenuse adjacent θ angle

Example 1: $\sqrt{a^2 - x^2}$ form

Evaluate

$$\int \sqrt{9 - x^2}\,dx$$

Here $a = 3$. Let $x = 3\sin\theta$, so $dx = 3\cos\theta\,d\theta$.

$$\sqrt{9 - x^2} = \sqrt{9 - 9\sin^2\theta} = 3\cos\theta$$

The integral becomes

$$\int 3\cos\theta \cdot 3\cos\theta\,d\theta = 9\int \cos^2\theta\,d\theta$$

Use the half-angle identity $\cos^2\theta = (1 + \cos 2\theta)/2$

$$9\int \frac{1 + \cos 2\theta}{2}\,d\theta = \frac{9}{2}\left(\theta + \frac{\sin 2\theta}{2}\right) + C$$

Since $\sin(2\theta) = 2\sin\theta\cos\theta$

$$= \frac{9}{2}\theta + \frac{9}{2}\sin\theta\cos\theta + C$$

Now convert back. From $x = 3\sin\theta$: $\sin\theta = x/3$, so $\theta = \arcsin(x/3)$. From the reference triangle: $\cos\theta = \sqrt{9 - x^2}/3$.

$$= \frac{9}{2}\arcsin\!\left(\frac{x}{3}\right) + \frac{x\sqrt{9 - x^2}}{2} + C$$

Example 2: $\sqrt{x^2 + a^2}$ form

Evaluate

$$\int \frac{dx}{\sqrt{x^2 + 4}}$$

Here $a = 2$. Let $x = 2\tan\theta$, so $dx = 2\sec^2\theta\,d\theta$.

$$\sqrt{x^2 + 4} = \sqrt{4\tan^2\theta + 4} = 2\sec\theta$$

The integral becomes

$$\int \frac{2\sec^2\theta\,d\theta}{2\sec\theta} = \int \sec\theta\,d\theta = \ln|\sec\theta + \tan\theta| + C$$

From the reference triangle with $x = 2\tan\theta$: $\tan\theta = x/2$, and the hypotenuse is $\sqrt{x^2 + 4}$, so $\sec\theta = \sqrt{x^2 + 4}/2$.

$$= \ln\left|\frac{\sqrt{x^2+4}}{2} + \frac{x}{2}\right| + C = \ln\left|x + \sqrt{x^2+4}\right| + C_1$$

(The ln(2) got absorbed into the constant.)

Example 3: $\sqrt{x^2 - a^2}$ form

Evaluate

$$\int \frac{dx}{\sqrt{x^2 - 9}} \qquad (x > 3)$$

Here $a = 3$. Let $x = 3\sec\theta$, so $dx = 3\sec\theta\tan\theta\,d\theta$.

$$\sqrt{x^2 - 9} = \sqrt{9\sec^2\theta - 9} = 3\tan\theta$$

The integral becomes

$$\int \frac{3\sec\theta\tan\theta\,d\theta}{3\tan\theta} = \int \sec\theta\,d\theta = \ln|\sec\theta + \tan\theta| + C$$

From the reference triangle: $\sec\theta = x/3$, $\tan\theta = \sqrt{x^2 - 9}/3$.

$$= \ln\left|\frac{x}{3} + \frac{\sqrt{x^2-9}}{3}\right| + C = \ln\left|x + \sqrt{x^2-9}\right| + C_1$$

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 $\sqrt{a^2 - x^2}$
• 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 $\sqrt{1 + (dy/dx)^2}$, which is the $\sqrt{x^2 + a^2}$ pattern
• In engineering, stress analysis on curved beams and fluid flow around cylindrical objects use these integrals