# Trig Substitution Practice Problems with Answers

> 12 trigonometric substitution practice problems with answers and full solutions for √(a²−x²), √(a²+x²), √(x²−a²) and completing the square.

Source: https://practicecalculus.com/practice-problems/trig-substitution
Updated: 2026-11-06

$$ \sqrt{a^2 - x^2}:\ x = a\sin\theta \quad \sqrt{a^2 + x^2}:\ x = a\tan\theta \quad \sqrt{x^2 - a^2}:\ x = a\sec\theta $$

## At a glance

- **Problems:** 12, with answers and full solutions
- **Level:** Calculus 2
- **You need:** [Trig integrals](/practice-problems/trig-integrals) and the Pythagorean identities
- **Main idea:** Choose $x$ so that the expression under the root becomes a perfect square of a trig function
- **Practice with new numbers:** [Trig substitution lesson](/learn/calculus-2/integration-techniques/trig-substitution)

## Key takeaways

- Match the form: $a^2 - x^2$ uses $x = a\sin\theta$, $a^2 + x^2$ uses $x = a\tan\theta$, and $x^2 - a^2$ uses $x = a\sec\theta$.
- Replace $dx$ too: $x = a\sin\theta$ gives $dx = a\cos\theta\,d\theta$.
- Draw a right triangle from the substitution to switch back from $\theta$ to $x$.

These **trig substitution practice problems** cover all three standard forms, a few direct formula cases, definite integrals and completing the square. Try each one first. The answers come first so you can check quickly, then the full solutions.

Pick the substitution from the form under the root:

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

## Problems

**Formula level**

1. $\displaystyle\int\frac{1}{\sqrt{4 - x^2}}\,dx$
2. $\displaystyle\int\frac{1}{9 + x^2}\,dx$
3. $\displaystyle\int\frac{1}{\sqrt{x^2 + 9}}\,dx$

**Full substitutions**

4. $\displaystyle\int\sqrt{1 - x^2}\,dx$
5. $\displaystyle\int\frac{x^2}{\sqrt{9 - x^2}}\,dx$
6. $\displaystyle\int\frac{1}{x^2\sqrt{4 - x^2}}\,dx$
7. $\displaystyle\int\frac{1}{(x^2 + 4)^{3/2}}\,dx$
8. $\displaystyle\int\frac{\sqrt{x^2 - 1}}{x}\,dx$ for $x > 1$

**Definite integrals**

9. $\displaystyle\int_0^1\sqrt{1 - x^2}\,dx$
10. $\displaystyle\int_0^2\frac{1}{x^2 + 4}\,dx$

**Complete the square first**

11. $\displaystyle\int\frac{1}{x^2 + 2x + 5}\,dx$
12. $\displaystyle\int\frac{1}{\sqrt{3 - 2x - x^2}}\,dx$

## Answers

| Problem | Answer |
| --- | --- |
| 1 | $\arcsin\dfrac{x}{2} + C$ |
| 2 | $\dfrac{1}{3}\arctan\dfrac{x}{3} + C$ |
| 3 | $\ln\left\lvert x + \sqrt{x^2 + 9}\right\rvert + C$ |
| 4 | $\dfrac{1}{2}\left(\arcsin x + x\sqrt{1 - x^2}\right) + C$ |
| 5 | $\dfrac{9}{2}\arcsin\dfrac{x}{3} - \dfrac{x}{2}\sqrt{9 - x^2} + C$ |
| 6 | $-\dfrac{\sqrt{4 - x^2}}{4x} + C$ |
| 7 | $\dfrac{x}{4\sqrt{x^2 + 4}} + C$ |
| 8 | $\sqrt{x^2 - 1} - \operatorname{arcsec} x + C$ |
| 9 | $\dfrac{\pi}{4}$ |
| 10 | $\dfrac{\pi}{8}$ |
| 11 | $\dfrac{1}{2}\arctan\dfrac{x + 1}{2} + C$ |
| 12 | $\arcsin\dfrac{x + 1}{2} + C$ |

## Step-by-step solutions

**1.** Let $x = 2\sin\theta$, so $dx = 2\cos\theta\,d\theta$ and $\sqrt{4 - x^2} = 2\cos\theta$. The integral becomes $\int d\theta = \theta$:

$$\arcsin\frac{x}{2} + C$$

**2.** Let $x = 3\tan\theta$, so $dx = 3\sec^2\theta\,d\theta$ and $9 + x^2 = 9\sec^2\theta$:

$$\int\frac{3\sec^2\theta}{9\sec^2\theta}\,d\theta = \frac{\theta}{3} + C = \frac{1}{3}\arctan\frac{x}{3} + C$$

**3.** Let $x = 3\tan\theta$, so $\sqrt{x^2 + 9} = 3\sec\theta$ and the integral becomes $\int\sec\theta\,d\theta = \ln|\sec\theta + \tan\theta|$. From the triangle, $\sec\theta = \frac{\sqrt{x^2+9}}{3}$ and $\tan\theta = \frac{x}{3}$. The $\ln 3$ that comes out joins the constant:

$$\ln\left|x + \sqrt{x^2 + 9}\right| + C$$

**4.** Let $x = \sin\theta$, so the integral is $\int\cos^2\theta\,d\theta = \frac{\theta}{2} + \frac{\sin\theta\cos\theta}{2}$. With $\sin\theta = x$ and $\cos\theta = \sqrt{1 - x^2}$, you get the answer in the table. See the [integral of √(1 − x²)](/integrals/sqrt-1-minus-x-squared).

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

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

Since $\sin\theta\cos\theta = \frac{x}{3}\cdot\frac{\sqrt{9 - x^2}}{3}$, this gives $\frac{9}{2}\arcsin\frac{x}{3} - \frac{x}{2}\sqrt{9 - x^2} + C$.

**6.** Let $x = 2\sin\theta$. Then $x^2 = 4\sin^2\theta$ and $\sqrt{4 - x^2} = 2\cos\theta$:

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

From the triangle, $\cot\theta = \frac{\sqrt{4 - x^2}}{x}$.

**7.** Let $x = 2\tan\theta$. Then $(x^2 + 4)^{3/2} = 8\sec^3\theta$ and $dx = 2\sec^2\theta\,d\theta$:

$$\int\frac{2\sec^2\theta}{8\sec^3\theta}\,d\theta = \frac{1}{4}\int\cos\theta\,d\theta = \frac{\sin\theta}{4} + C$$

From the triangle, $\sin\theta = \frac{x}{\sqrt{x^2 + 4}}$.

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

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

With $\tan\theta = \sqrt{x^2 - 1}$ and $\theta = \operatorname{arcsec} x$, you get the answer.

**9.** Use problem 4: $\frac{1}{2}\left(\arcsin 1 + 0\right) - 0 = \frac{\pi}{4}$. This is a quarter of the unit circle.

**10.** Use $\frac{1}{2}\arctan\frac{x}{2}$:

$$\frac{1}{2}\arctan 1 - \frac{1}{2}\arctan 0 = \frac{\pi}{8}$$

**11.** Complete the square: $x^2 + 2x + 5 = (x + 1)^2 + 4$. With $u = x + 1$, this is $\int\frac{du}{u^2 + 4} = \frac{1}{2}\arctan\frac{u}{2}$.

**12.** Complete the square: $3 - 2x - x^2 = 4 - (x + 1)^2$. With $u = x + 1$, this is $\int\frac{du}{\sqrt{4 - u^2}} = \arcsin\frac{u}{2}$.

## Common mistakes on these problems

- **Choosing the wrong substitution.** $\sqrt{9 + x^2}$ needs $\tan$, not $\sin$. Match the sign pattern in the table above.
- **Forgetting to replace $dx$.** With $x = 2\tan\theta$, $dx = 2\sec^2\theta\,d\theta$. Leaving $dx$ alone breaks every later step.
- **Leaving the answer in $\theta$.** An indefinite integral in $x$ needs an answer in $x$. Use the triangle.
- **Using trig substitution when u-substitution works.** $\int x\sqrt{4 - x^2}\,dx$ is one line with $u = 4 - x^2$.

## Keep practicing

For unlimited problems with instant feedback, try the [trig substitution lesson](/learn/calculus-2/integration-techniques/trig-substitution) and the [choosing a strategy lesson](/learn/calculus-2/integration-techniques/choosing-a-strategy). For worked examples, read the [trig substitution guide](/blog/trig-substitution). For rational functions without roots, move on to [partial fractions problems](/practice-problems/partial-fractions).

## Frequently asked questions

### When should you use trig substitution?

Use it when you see $\sqrt{a^2 - x^2}$, $\sqrt{a^2 + x^2}$, $\sqrt{x^2 - a^2}$, or $a^2 + x^2$ raised to a power, and a simpler u-substitution does not work. If an $x$ factor sits outside the root, as in $\int x\sqrt{1 - x^2}\,dx$, try u-substitution first.

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

Draw a right triangle. For $x = 2\sin\theta$, label the opposite side $x$ and the hypotenuse $2$. The adjacent side is then $\sqrt{4 - x^2}$, and you can read off $\cos\theta$, $\tan\theta$ or any other ratio.

### What do you do with a quadratic that is not in the standard form?

Complete the square first. For example, $x^2 + 2x + 5 = (x + 1)^2 + 4$, which matches $u^2 + a^2$ with $u = x + 1$ and $a = 2$. Problems 11 and 12 show this.