Integration by Trigonometric Substitution Tool

Math Calculus • Applications of Integrals

Written by STEM Calculators Team Published December 28, 2025 Updated February 24, 2026

10. Integration By Trigonometric Substitution Tool

Applies trig substitutions for radicals: \(\;\sqrt{a^2-x^2}\Rightarrow x=a\sin\theta,\quad \sqrt{x^2+a^2}\Rightarrow x=a\tan\theta,\quad \sqrt{x^2-a^2}\Rightarrow x=a\sec\theta\;\) and back-substitutes using a right-triangle diagram.

Inputs

Integrand \(f(x)\)

Use variable x. Constants: pi, e, and a (from the field below). Supported: + − * / ^, parentheses, sin cos tan ln log sqrt abs exp. Implicit multiplication allowed: 2x, (x+1)(x-1).

Parameter \(a\)

Used when your expression contains a.

Substitution type

Auto works for the classic three patterns.

Triangle example \(x\)

Used only to label the triangle sides.

Angle \(\theta\) (deg)

Used for \(\theta\to x\) preview (optional).

Options

Show grid Show steps

Presets

Click a preset to load and evaluate.

Ready

Visuals

Graph: drag to pan • wheel/pinch to zoom. Triangle: conversion between \(x\) and \(\theta\).

Graph

x: 0, y: 0, zoom(px/unit): 60

Right triangle diagram

\(\theta\) is drawn at the acute angle (left corner), not at the right angle.

Result & Steps

Enter an integrand, then click Calculate.

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10. Integration By Trigonometric Substitution Tool — Theory

When trigonometric substitution is used

Trig substitution is designed for integrals containing radicals like:

\[ \sqrt{a^2-x^2},\qquad \sqrt{x^2+a^2},\qquad \sqrt{x^2-a^2}. \]

The goal is to rewrite the radical using a trig identity so the square root disappears and the integral becomes an integral in \(\theta\).

The three standard substitutions

\[ \begin{aligned} \sqrt{a^2-x^2}:&\quad x=a\sin\theta,\ \ dx=a\cos\theta\,d\theta,\ \ \sqrt{a^2-x^2}=a\cos\theta.\\[4pt] \sqrt{x^2+a^2}:&\quad x=a\tan\theta,\ \ dx=a\sec^2\theta\,d\theta,\ \ \sqrt{x^2+a^2}=a\sec\theta.\\[4pt] \sqrt{x^2-a^2}:&\quad x=a\sec\theta,\ \ dx=a\sec\theta\tan\theta\,d\theta,\ \ \sqrt{x^2-a^2}=a\tan\theta. \end{aligned} \]

The identities behind these are:

\[ 1-\sin^2\theta=\cos^2\theta,\qquad 1+\tan^2\theta=\sec^2\theta,\qquad \sec^2\theta-1=\tan^2\theta. \]

Right-triangle back-substitution

After integrating in \(\theta\), you must rewrite the answer back in terms of \(x\). This is done by building a right triangle that matches the defining ratio.

\[ \sin\theta=\frac{x}{a}\Rightarrow \cos\theta=\frac{\sqrt{a^2-x^2}}{a},\ \ \theta=\arcsin\!\left(\frac{x}{a}\right) \]

\[ \tan\theta=\frac{x}{a}\Rightarrow \sec\theta=\frac{\sqrt{x^2+a^2}}{a},\ \ \theta=\arctan\!\left(\frac{x}{a}\right) \]

\[ \sec\theta=\frac{x}{a}\Rightarrow \tan\theta=\frac{\sqrt{x^2-a^2}}{a},\ \ \theta=\operatorname{arcsec}\!\left(\frac{x}{a}\right)=\arccos\!\left(\frac{a}{x}\right) \]

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

Here \(a=3\). Use \(x=3\sin\theta\), so \(dx=3\cos\theta\,d\theta\) and \(\sqrt{9-x^2}=\sqrt{9-9\sin^2\theta}=3\cos\theta\).

\[ \int \sqrt{9-x^2}\,dx = \int (3\cos\theta)(3\cos\theta)\,d\theta = 9\int \cos^2\theta\,d\theta. \]

Use \(\cos^2\theta=\frac{1+\cos2\theta}{2}\) to integrate:

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

Back-substitute using \(\theta=\arcsin(x/3)\), \(\sin\theta=x/3\), \(\cos\theta=\sqrt{9-x^2}/3\):

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

University note: hyperbolic substitutions

Some integrals with \(\sqrt{x^2\pm a^2}\) can also be simplified using hyperbolic substitutions (e.g., \(x=a\sinh u\), \(x=a\cosh u\)), but trig substitution is the standard method taught first and is what this tool demonstrates.

Frequently Asked Questions

What is integration by trigonometric substitution?

It is a method that replaces x with a trig expression so radicals simplify using identities like 1 - sin^2(theta) = cos^2(theta) or 1 + tan^2(theta) = sec^2(theta). This turns an integral with a square root into a simpler integral in theta.

When should I use x = a sin(theta) for sqrt(a^2 - x^2)?

Use x = a sin(theta) when your integrand contains sqrt(a^2 - x^2) because sqrt(a^2 - a^2 sin^2(theta)) becomes a cos(theta). This removes the square root and makes the integral easier to evaluate.

How does the auto-detect option choose the substitution?

Auto-detect looks for the standard radical forms sqrt(a^2 - x^2), sqrt(x^2 + a^2), or sqrt(x^2 - a^2) and selects the corresponding substitution. If your integrand does not match a classic pattern, you may need to choose the substitution manually.

How do I convert the answer back from theta to x?

After integrating in theta, use the defining ratio (such as sin(theta) = x/a or tan(theta) = x/a) to build a right triangle and rewrite trig terms in x. Then replace theta with an inverse trig expression like theta = arcsin(x/a) or theta = arctan(x/a).

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