Trigonometric Substitution Tool

Math Calculus • Integrals

Written by STEM Calculators Team Published June 15, 2026 Updated June 15, 2026

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Theory

What is trigonometric substitution?

Trigonometric substitution is a method for integrals involving square-root expressions such as:

\[ \begin{aligned} \sqrt{a^2-x^2},\qquad \sqrt{x^2-a^2},\qquad \sqrt{x^2+a^2} \end{aligned} \]

The goal is to replace \(x\) with a trigonometric expression so that the square root simplifies by a Pythagorean identity.

The substitution is chosen from the shape of the expression under the square root.

The three standard substitutions

The three most important forms are:

\[ \begin{aligned} \sqrt{a^2-x^2} &\quad\Rightarrow\quad x=a\sin\theta\\[4pt] \sqrt{x^2-a^2} &\quad\Rightarrow\quad x=a\sec\theta\\[4pt] \sqrt{x^2+a^2} &\quad\Rightarrow\quad x=a\tan\theta \end{aligned} \]

These choices are based on the identities:

\[ \begin{aligned} 1-\sin^2\theta&=\cos^2\theta\\ \sec^2\theta-1&=\tan^2\theta\\ 1+\tan^2\theta&=\sec^2\theta \end{aligned} \]

Form 1: \(\sqrt{a^2-x^2}\)

For:

\[ \begin{aligned} \sqrt{a^2-x^2} \end{aligned} \]

use:

\[ \begin{aligned} x&=a\sin\theta\\ dx&=a\cos\theta\,d\theta \end{aligned} \]

Then:

\[ \begin{aligned} \sqrt{a^2-x^2} &= \sqrt{a^2-a^2\sin^2\theta}\\ &= a\sqrt{1-\sin^2\theta}\\ &= a\cos\theta \end{aligned} \]

The reference triangle has:

\[ \begin{aligned} \sin\theta&=\frac{x}{a}\\ \cos\theta&=\frac{\sqrt{a^2-x^2}}{a} \end{aligned} \]

Worked example: \(\int \frac{dx}{\sqrt{9-x^2}}\)

Start with:

\[ \begin{aligned} \int \frac{dx}{\sqrt{9-x^2}} \end{aligned} \]

Here:

\[ \begin{aligned} a^2&=9\\ a&=3 \end{aligned} \]

Step 1. Use the substitution.

\[ \begin{aligned} x&=3\sin\theta\\ dx&=3\cos\theta\,d\theta \end{aligned} \]

Step 2. Simplify the radical.

\[ \begin{aligned} \sqrt{9-x^2} &= \sqrt{9-9\sin^2\theta}\\ &= 3\cos\theta \end{aligned} \]

Step 3. Substitute into the integral.

\[ \begin{aligned} \int \frac{dx}{\sqrt{9-x^2}} &= \int \frac{3\cos\theta\,d\theta}{3\cos\theta}\\ &= \int d\theta\\ &= \theta+C \end{aligned} \]

Step 4. Convert back to \(x\).

Since \(x=3\sin\theta\), we have:

\[ \begin{aligned} \sin\theta&=\frac{x}{3}\\ \theta&=\arcsin\!\left(\frac{x}{3}\right) \end{aligned} \]

Therefore:

\[ \begin{aligned} \int \frac{dx}{\sqrt{9-x^2}} &= \arcsin\!\left(\frac{x}{3}\right)+C \end{aligned} \]

Form 2: \(\sqrt{x^2-a^2}\)

For:

\[ \begin{aligned} \sqrt{x^2-a^2} \end{aligned} \]

use:

\[ \begin{aligned} x&=a\sec\theta\\ dx&=a\sec\theta\tan\theta\,d\theta \end{aligned} \]

Then:

\[ \begin{aligned} \sqrt{x^2-a^2} &= \sqrt{a^2\sec^2\theta-a^2}\\ &= a\sqrt{\sec^2\theta-1}\\ &= a\tan\theta \end{aligned} \]

The reference triangle has:

\[ \begin{aligned} \sec\theta&=\frac{x}{a}\\ \tan\theta&=\frac{\sqrt{x^2-a^2}}{a} \end{aligned} \]

A common result is:

\[ \begin{aligned} \int \frac{dx}{\sqrt{x^2-a^2}} &= \ln\left|x+\sqrt{x^2-a^2}\right|+C \end{aligned} \]

Form 3: \(\sqrt{x^2+a^2}\)

For:

\[ \begin{aligned} \sqrt{x^2+a^2} \end{aligned} \]

use:

\[ \begin{aligned} x&=a\tan\theta\\ dx&=a\sec^2\theta\,d\theta \end{aligned} \]

Then:

\[ \begin{aligned} \sqrt{x^2+a^2} &= \sqrt{a^2\tan^2\theta+a^2}\\ &= a\sqrt{\tan^2\theta+1}\\ &= a\sec\theta \end{aligned} \]

The reference triangle has:

\[ \begin{aligned} \tan\theta&=\frac{x}{a}\\ \sec\theta&=\frac{\sqrt{x^2+a^2}}{a} \end{aligned} \]

A common result is:

\[ \begin{aligned} \int \frac{dx}{\sqrt{x^2+a^2}} &= \ln\left|x+\sqrt{x^2+a^2}\right|+C \end{aligned} \]

Square-root integrals themselves

Trig substitution also handles integrals where the entire square-root expression is in the numerator.

Important formulas are:

\[ \begin{aligned} \int \sqrt{a^2-x^2}\,dx &= \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\arcsin\!\left(\frac{x}{a}\right)+C\\[6pt] \int \sqrt{x^2-a^2}\,dx &= \frac{x}{2}\sqrt{x^2-a^2} - \frac{a^2}{2}\ln\left|x+\sqrt{x^2-a^2}\right|+C\\[6pt] \int \sqrt{x^2+a^2}\,dx &= \frac{x}{2}\sqrt{x^2+a^2} + \frac{a^2}{2}\ln\left|x+\sqrt{x^2+a^2}\right|+C \end{aligned} \]

Why the right triangle matters

After integrating in terms of \(\theta\), the answer must be written again in terms of \(x\). The reference triangle gives the conversion.

For example, if:

\[ \begin{aligned} x&=a\sin\theta \end{aligned} \]

then:

\[ \begin{aligned} \sin\theta&=\frac{x}{a} \end{aligned} \]

So:

\[ \begin{aligned} \theta&=\arcsin\!\left(\frac{x}{a}\right) \end{aligned} \]

The triangle also gives expressions for \(\cos\theta\), \(\tan\theta\), and \(\sec\theta\) in terms of \(x\).

Definite integrals and domains

After finding an antiderivative \(F(x)\), a definite integral is evaluated by:

\[ \begin{aligned} \int_{a_0}^{b_0} f(x)\,dx &= F(b_0)-F(a_0) \end{aligned} \]

However, the real domain matters:

\(\sqrt{a^2-x^2}\) requires \(-a\le x\le a\).
\(\sqrt{x^2-a^2}\) requires \(x\le -a\) or \(x\ge a\).
\(\sqrt{x^2+a^2}\) is real for all \(x\).

If a selected interval leaves the real domain, the real-valued definite integral is not valid on that whole interval.

Graph interpretation

The graph shows:

the integrand \(f(x)\),
the antiderivative \(F(x)+C\),
the shaded interval \([a_0,b_0]\),
domain boundary lines such as \(x=\pm a\), when relevant.

The shaded region represents:

\[ \begin{aligned} \int_{a_0}^{b_0} f(x)\,dx \end{aligned} \]

Play mode moves across the interval and shows the accumulated area.

Common mistakes

Choosing the wrong substitution: match the form under the square root first.
Forgetting \(dx\): after substituting \(x\), always compute \(dx\).
Ignoring the domain: square roots restrict where the integrand is real.
Forgetting the triangle: the triangle is needed to convert back from \(\theta\) to \(x\).
Missing constants: if \(a=3\), then \(a^2=9\), and the substitution must use \(x=3\sin\theta\), \(x=3\sec\theta\), or \(x=3\tan\theta\).
Forgetting \(+C\): every indefinite integral needs the arbitrary constant.

Frequently Asked Questions

Which substitution is used for sqrt(a^2 - x^2)?

Use x = a sin(theta). Then sqrt(a^2 - x^2) becomes a cos(theta).

Which substitution is used for sqrt(x^2 - a^2)?

Use x = a sec(theta). Then sqrt(x^2 - a^2) becomes a tan(theta).

Which substitution is used for sqrt(x^2 + a^2)?

Use x = a tan(theta). Then sqrt(x^2 + a^2) becomes a sec(theta).

What is the integral of 1 over sqrt(9 minus x squared)?

The result is arcsin(x/3) plus C.

Why is a right triangle useful?

The triangle helps convert trigonometric expressions in theta back into expressions involving x.

What happens if my interval is outside the real domain?

The calculator warns that the selected interval leaves the real domain, and the definite integral may be undefined as a real-valued integral.

Can trig substitution produce logarithms?

Yes. The sqrt(x^2 - a^2) and sqrt(x^2 + a^2) forms often produce logarithmic antiderivatives.

What does the shaded graph region represent?

The shaded region represents the signed definite integral of the selected integrand over the selected interval.

Trigonometric Substitution Tool

Integral and substitution form

Graph and output settings

Quick examples

Trigonometric substitution graph

Step-by-step trigonometric substitution

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Frequently Asked Questions

Integral and substitution form

Graph and output settings

Quick examples

Trigonometric substitution graph

Step-by-step trigonometric substitution

Rate this calculator

Frequently Asked Questions

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