# Trig Substitution: The Three Cases, Triangles and Examples

> Learn trigonometric substitution: which of the three substitutions to use, how to build the reference triangle, and how to convert the answer back to x.

Source: https://practicecalculus.com/blog/trig-substitution
Updated: 2026-09-06

## Key takeaways

- Use $x = a\sin\theta$ for $\sqrt{a^2 - x^2}$, $x = a\tan\theta$ for $\sqrt{a^2 + x^2}$, and $x = a\sec\theta$ for $\sqrt{x^2 - a^2}$.
- Each substitution turns the square root into a single trig function using a Pythagorean identity.
- A right triangle built from the substitution converts any trig expression in $\theta$ back into $x$.
- Try u-substitution first: if the derivative of the inside of the root is present, trig substitution is overkill.

The integral $\int \sqrt{4 - x^2}\,dx$ looks innocent, yet u-substitution fails (there is no $x$ outside the root to absorb $du$) and integration by parts goes nowhere. **Trig substitution** removes the square root entirely by replacing $x$ with a trig function, and it is the standard way to integrate expressions involving $\sqrt{a^2 - x^2}$, $\sqrt{a^2 + x^2}$ and $\sqrt{x^2 - a^2}$.

## What is trig substitution?

Trigonometric substitution replaces $x$ with $a\sin\theta$, $a\tan\theta$ or $a\sec\theta$ so that a Pythagorean identity collapses the square root into a single term. For example, with $x = 2\sin\theta$:

$$\sqrt{4 - x^2} = \sqrt{4 - 4\sin^2\theta} = \sqrt{4\cos^2\theta} = 2\cos\theta$$

The root is gone, and you are left with a trig integral. At the end you use a right triangle to rewrite the answer in terms of $x$.

## The three trig substitutions

| Expression        | Substitution         | $dx$                              | Identity used                       | Root becomes   |
| ----------------- | -------------------- | --------------------------------- | ----------------------------------- | -------------- |
| $\sqrt{a^2-x^2}$  | $x = a\sin\theta$    | $a\cos\theta\,d\theta$            | $1-\sin^2\theta = \cos^2\theta$     | $a\cos\theta$  |
| $\sqrt{a^2+x^2}$  | $x = a\tan\theta$    | $a\sec^2\theta\,d\theta$          | $1+\tan^2\theta = \sec^2\theta$     | $a\sec\theta$  |
| $\sqrt{x^2-a^2}$  | $x = a\sec\theta$    | $a\sec\theta\tan\theta\,d\theta$  | $\sec^2\theta-1 = \tan^2\theta$     | $a\tan\theta$  |

A quick way to remember: **minus $x^2$ means sine, plus means tangent, and $x^2$ minus means secant.**

The substitutions come with restricted angle ranges ($-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$ for sine, $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$ for tangent, and $0 \le \theta < \frac{\pi}{2}$ for secant when $x \ge a$). On these ranges $\cos\theta$, $\sec\theta$ and $\tan\theta$ are nonnegative, which is why the square roots simplify without absolute values.

## The reference triangles, described in words

To convert back to $x$, draw a right triangle with angle $\theta$ and label two sides straight from your substitution. The third side comes from the Pythagorean theorem, and it is always the square root from the problem.

- **Sine case, $x = a\sin\theta$.** Since $\sin\theta = \frac{x}{a} = \frac{\text{opposite}}{\text{hypotenuse}}$, put $x$ on the side opposite $\theta$ and $a$ on the hypotenuse. The adjacent side is $\sqrt{a^2 - x^2}$.
- **Tangent case, $x = a\tan\theta$.** Since $\tan\theta = \frac{x}{a} = \frac{\text{opposite}}{\text{adjacent}}$, put $x$ opposite and $a$ adjacent. The hypotenuse is $\sqrt{a^2 + x^2}$.
- **Secant case, $x = a\sec\theta$.** Since $\sec\theta = \frac{x}{a} = \frac{\text{hypotenuse}}{\text{adjacent}}$, put $x$ on the hypotenuse and $a$ adjacent. The opposite side is $\sqrt{x^2 - a^2}$.

Once the triangle is labeled, any trig function of $\theta$ is a ratio of two sides. For instance, in the sine triangle $\cot\theta = \frac{\text{adjacent}}{\text{opposite}} = \frac{\sqrt{a^2-x^2}}{x}$.

## The trig substitution method in five steps

1. **Identify the form** and choose the substitution from the table.
2. **Substitute** for $x$, $dx$ and the root.
3. **Simplify** and integrate the trig integral in $\theta$.
4. **Build the triangle** from your substitution.
5. **Convert back** to $x$ using the triangle (and $\theta = \arcsin\frac{x}{a}$ or similar for any bare $\theta$).

## Worked example 1: the sine substitution

Evaluate $\int \frac{dx}{x^2\sqrt{9-x^2}}$.

**Step 1. Identify the form.** $\sqrt{9 - x^2}$ is $\sqrt{a^2 - x^2}$ with $a = 3$, so let $x = 3\sin\theta$.

**Step 2. Substitute.** $dx = 3\cos\theta\,d\theta$, $x^2 = 9\sin^2\theta$ and $\sqrt{9 - x^2} = 3\cos\theta$:

$$\int \frac{3\cos\theta\,d\theta}{9\sin^2\theta\cdot 3\cos\theta} = \frac19\int \frac{d\theta}{\sin^2\theta} = \frac19\int \csc^2\theta\,d\theta$$

**Step 3. Integrate.** $\int\csc^2\theta\,d\theta = -\cot\theta$, so the result is $-\frac19\cot\theta + C$.

**Step 4. Build the triangle.** $\sin\theta = \frac{x}{3}$: opposite $x$, hypotenuse $3$, adjacent $\sqrt{9 - x^2}$.

**Step 5. Convert back.** $\cot\theta = \frac{\sqrt{9-x^2}}{x}$, so

$$\int \frac{dx}{x^2\sqrt{9-x^2}} = -\frac{\sqrt{9-x^2}}{9x} + C$$

## Worked example 2: area of a quarter circle

Evaluate $\int_0^2 \sqrt{4 - x^2}\,dx$.

The curve $y = \sqrt{4 - x^2}$ is the top half of the circle of radius 2, so this integral should equal a quarter of the circle's area, $\frac14\pi(2)^2 = \pi$. Let's confirm it.

**Step 1.** Let $x = 2\sin\theta$, so $dx = 2\cos\theta\,d\theta$ and $\sqrt{4 - x^2} = 2\cos\theta$.

**Step 2. Change the bounds.** When $x = 0$, $\sin\theta = 0$, so $\theta = 0$. When $x = 2$, $\sin\theta = 1$, so $\theta = \frac{\pi}{2}$.

**Step 3. Substitute.**

$$\int_0^{\pi/2} 2\cos\theta\cdot 2\cos\theta\,d\theta = 4\int_0^{\pi/2}\cos^2\theta\,d\theta$$

**Step 4. Use the half-angle identity** $\cos^2\theta = \frac{1 + \cos 2\theta}{2}$:

$$4\int_0^{\pi/2}\frac{1 + \cos 2\theta}{2}\,d\theta = 2\left[\theta + \tfrac12\sin 2\theta\right]_0^{\pi/2} = 2\left(\tfrac{\pi}{2} + 0\right) - 0 = \pi$$

Because we changed the bounds, there was no need to convert back to $x$.

```graph
title: 'Quarter circle under y = √(4 − x²) from x = 0 to x = 2'
caption: 'The shaded region is a quarter of a circle of radius 2, and trig substitution confirms $\int_0^2 \sqrt{4-x^2}\,dx = \pi$.'
x: [-2.5, 2.5]
y: [-0.5, 2.5]
functions:
  - expr: sqrt(4-x^2)
    label: y = √(4 − x²)
shade: {expr: sqrt(4-x^2), from: 0, to: 2}
points:
  - [0, 2, '(0, 2)']
  - [2, 0, '(2, 0)']
```

## Worked example 3: the tangent substitution

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

**Step 1.** The form is $\sqrt{a^2 + x^2}$ with $a = 2$. Let $x = 2\tan\theta$.

**Step 2.** $dx = 2\sec^2\theta\,d\theta$ and $\sqrt{x^2 + 4} = \sqrt{4\tan^2\theta + 4} = 2\sec\theta$:

$$\int \frac{2\sec^2\theta\,d\theta}{2\sec\theta} = \int \sec\theta\,d\theta$$

**Step 3.** $\int\sec\theta\,d\theta = \ln|\sec\theta + \tan\theta| + C$.

**Step 4. Triangle.** $\tan\theta = \frac{x}{2}$: opposite $x$, adjacent $2$, hypotenuse $\sqrt{x^2 + 4}$. So $\sec\theta = \frac{\sqrt{x^2+4}}{2}$.

**Step 5. Convert back.**

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

The $-\ln 2$ is a constant, so it merges into $C$:

$$\int \frac{dx}{\sqrt{x^2+4}} = \ln\left(x + \sqrt{x^2+4}\right) + C$$

## Worked example 4: the secant substitution

Evaluate $\int \frac{\sqrt{x^2-1}}{x}\,dx$ for $x > 1$.

**Step 1.** The form is $\sqrt{x^2 - a^2}$ with $a = 1$. Let $x = \sec\theta$ with $0 \le \theta < \frac{\pi}{2}$.

**Step 2.** $dx = \sec\theta\tan\theta\,d\theta$ and $\sqrt{x^2 - 1} = \tan\theta$:

$$\int \frac{\tan\theta}{\sec\theta}\cdot\sec\theta\tan\theta\,d\theta = \int\tan^2\theta\,d\theta$$

**Step 3.** Use $\tan^2\theta = \sec^2\theta - 1$:

$$\int(\sec^2\theta - 1)\,d\theta = \tan\theta - \theta + C$$

**Step 4. Triangle.** $\sec\theta = x$: hypotenuse $x$, adjacent $1$, opposite $\sqrt{x^2-1}$. So $\tan\theta = \sqrt{x^2-1}$, and $\theta = \operatorname{arcsec} x$.

**Step 5.**

$$\int \frac{\sqrt{x^2-1}}{x}\,dx = \sqrt{x^2-1} - \operatorname{arcsec} x + C$$

## Worked example 5: converting a double angle back to x

Evaluate $\int \frac{x^2}{\sqrt{1-x^2}}\,dx$.

**Step 1.** The form is $\sqrt{a^2 - x^2}$ with $a = 1$. Let $x = \sin\theta$.

**Step 2.** $dx = \cos\theta\,d\theta$ and $\sqrt{1 - x^2} = \cos\theta$:

$$\int \frac{\sin^2\theta}{\cos\theta}\cdot\cos\theta\,d\theta = \int \sin^2\theta\,d\theta$$

**Step 3.** Use $\sin^2\theta = \frac{1 - \cos 2\theta}{2}$:

$$\int \frac{1 - \cos 2\theta}{2}\,d\theta = \frac{\theta}{2} - \frac{\sin 2\theta}{4} + C$$

**Step 4. Rewrite the double angle.** The triangle only gives you functions of $\theta$, not $2\theta$, so use $\sin 2\theta = 2\sin\theta\cos\theta$:

$$\frac{\theta}{2} - \frac{\sin\theta\cos\theta}{2} + C$$

**Step 5. Convert back.** The triangle for $\sin\theta = x$ has opposite $x$, hypotenuse $1$ and adjacent $\sqrt{1-x^2}$, so $\cos\theta = \sqrt{1 - x^2}$. And $\theta = \arcsin x$.

$$\int \frac{x^2}{\sqrt{1-x^2}}\,dx = \frac{\arcsin x}{2} - \frac{x\sqrt{1-x^2}}{2} + C$$

> **Tip:** whenever your $\theta$-answer contains $\sin 2\theta$ or $\cos 2\theta$, expand it with a double-angle identity before using the triangle.

## Completing the square first

If the root holds a general quadratic, complete the square to reach one of the three forms. For $\int \frac{dx}{\sqrt{x^2 + 2x + 5}}$:

$$x^2 + 2x + 5 = (x+1)^2 + 4$$

Let $u = x + 1$ (so $du = dx$). The integral becomes $\int \frac{du}{\sqrt{u^2 + 4}}$, which is exactly example 3:

$$\int \frac{dx}{\sqrt{x^2+2x+5}} = \ln\left(x + 1 + \sqrt{x^2+2x+5}\right) + C$$

## When not to use trig substitution

Always check for a simpler method first. In $\int \frac{x}{\sqrt{9 - x^2}}\,dx$, the factor $x$ is (up to a constant) the derivative of $9 - x^2$, so [u-substitution](/blog/u-substitution) with $u = 9 - x^2$ gives $-\sqrt{9 - x^2} + C$ in two lines. Trig substitution would work, but it is the long way around. Similarly, a rational function with no root is usually a job for [partial fractions](/blog/partial-fractions).

## Common trig substitution mistakes

1. **Picking the wrong substitution.** $\sqrt{x^2 - 9}$ needs secant, not sine. Match the sign pattern carefully.
2. **Forgetting to substitute $dx$.** $dx$ becomes $a\cos\theta\,d\theta$ (or the matching expression), not $d\theta$.
3. **Leaving the answer in $\theta$.** An indefinite integral must end in $x$. Use the triangle.
4. **Mislabeling the triangle.** Write the defining ratio ($\sin\theta = \frac{x}{a}$, for example) and place sides from it, rather than from memory.
5. **Using old bounds.** For definite integrals, convert the bounds to $\theta$ values, or convert back to $x$ before evaluating.

## Practice the three cases until they click

Trig substitution has a lot of moving parts, but every problem follows the same five steps. The fastest way to get comfortable is to do several problems in a row so the triangles become automatic. The interactive lesson below gives you instant feedback on each substitution and shows the triangle whenever you get stuck.

## Frequently asked questions

### When do you use trig substitution?

Use it when an integrand contains $\sqrt{a^2 - x^2}$, $\sqrt{a^2 + x^2}$ or $\sqrt{x^2 - a^2}$ (or those expressions raised to a power) and a simple u-substitution does not work. The substitution removes the square root using a Pythagorean identity.

### How do you know which trig substitution to use?

Match the form under the root. A constant minus $x^2$ uses sine, a constant plus $x^2$ uses tangent, and $x^2$ minus a constant uses secant. In each case the identity $1 - \sin^2\theta = \cos^2\theta$, $1 + \tan^2\theta = \sec^2\theta$ or $\sec^2\theta - 1 = \tan^2\theta$ does the work.

### How do you convert back to x after trig substitution?

Draw a right triangle that matches your substitution. For $x = a\sin\theta$, the opposite side is $x$, the hypotenuse is $a$, and the adjacent side is $\sqrt{a^2 - x^2}$ by the Pythagorean theorem. Then read off any trig function of $\theta$ as a ratio of sides.

### What if the expression under the root is not in the form a² ± x²?

Complete the square first. For example $x^2 + 2x + 5 = (x+1)^2 + 4$, so with $u = x + 1$ the root becomes $\sqrt{u^2 + 4}$ and you can use $u = 2\tan\theta$.

### Is trig substitution the same as u-substitution?

Both change variables, but in opposite directions. In u-substitution you set $u$ equal to a function of $x$. In trig substitution you set $x$ equal to a function of $\theta$, which is why you need a triangle to get back to $x$ at the end.