Some integrals refuse to yield to algebraic substitution, but the moment you swap $x$ for $a\sin\theta$, $a\tan\theta$ or $a\sec\theta$, the radical collapses by a Pythagorean identity and the integral falls out. This lesson introduces the three standard patterns, the geometry behind each choice, and the algebra of converting back from $\theta$ to $x$.
Today's hook, Try to integrate $\displaystyle\int \frac{1}{\sqrt{9-x^2}}\,dx$ using $u$-substitution. You will hit a wall, there is no inner derivative to cancel. Now try $x = 3\sin\theta$, so $dx = 3\cos\theta\,d\theta$. Watch the radical melt away. Compare your answer after Card 05.
0/5QUESTS
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You are here
Recall, your gut answer first
+5 XP warm-up Without integrating, decide which substitution will simplify each radical: (a) $\sqrt{16-x^2}$, (b) $\sqrt{x^2+25}$, (c) $\sqrt{x^2-4}$.
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Recall, your gut answer first
+5 XP warm-up
Without integrating, decide which substitution will simplify each radical: (a) $\sqrt{16-x^2}$, (b) $\sqrt{x^2+25}$, (c) $\sqrt{x^2-4}$. Before checking match each to one of $x=a\sin\theta$, $x=a\tan\theta$, $x=a\sec\theta$ and state the value of $a$.
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The two moves for trig substitution
+5 XP to read Every trig substitution reduces to two disciplined moves: match the radical to a Pythagorean identity ($1-\sin^2 = \cos^2$, $1+\tan^2 = \sec^2$, $\sec^2 - 1 = \tan^2$), then substitute and replace $dx$ using $dx = a\cos\theta\,d\theta$ (or its tan/sec analogue).
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The two moves for trig substitution
+5 XP to read
Every trig substitution reduces to two disciplined moves: match the radical to a Pythagorean identity ($1-\sin^2 = \cos^2$, $1+\tan^2 = \sec^2$, $\sec^2 - 1 = \tan^2$), then substitute and replace $dx$ using $dx = a\cos\theta\,d\theta$ (or its tan/sec analogue). Once the radical collapses, the integral becomes a standard trig integral.
The match-substitute-simplify routine: (1) identify the radical pattern, (2) substitute $x$ and compute $dx$, (3) use the matching identity to collapse the radical.
NESA outcome MEX-C1: applies techniques of integration
Understand
Concepts
Why each substitution collapses the radical via a Pythagorean identity
Why the principal range matters (so $\cos\theta \geq 0$, etc.)
How a right-triangle diagram converts back from $\theta$ to $x$
Can do
Skills
Choose the correct trig substitution from the form of the integrand
Carry out the full substitution including $dx$
Convert the answer back from $\theta$ to $x$ using a reference triangle
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Key terms
Trigonometric substitutionA change of variable replacing $x$ with a trig function of $\theta$ so a Pythagorean identity collapses a radical of the form $\sqrt{a^2 \pm x^2}$ or $\sqrt{x^2 - a^2}$.
Sine substitution$x = a\sin\theta$ with $\theta \in [-\pi/2, \pi/2]$. Collapses $\sqrt{a^2 - x^2}$ to $a\cos\theta$. Use when the radical is $\sqrt{a^2 - x^2}$.
Tangent substitution$x = a\tan\theta$ with $\theta \in (-\pi/2, \pi/2)$. Converts $a^2 + x^2$ to $a^2\sec^2\theta$. Use when the integrand contains $a^2 + x^2$ (no radical needed).
Secant substitution$x = a\sec\theta$ with $\theta \in [0, \pi/2)$. Collapses $\sqrt{x^2 - a^2}$ to $a\tan\theta$. Use when the radical is $\sqrt{x^2 - a^2}$ with $x \geq a$.
Reference triangleA right triangle drawn from the substitution (e.g., opposite $= x$, hypotenuse $= a$ for $\sin\theta = x/a$). Used to read off $\cos\theta$, $\tan\theta$, etc. in terms of $x$ at the end.
Pythagorean identityThe three forms $\sin^2\theta + \cos^2\theta = 1$, $1+\tan^2\theta = \sec^2\theta$, $\sec^2\theta - 1 = \tan^2\theta$, each matched to one substitution.
MEX-C1NESA outcome (Further Integration): applies techniques of integration including trigonometric substitution to evaluate integrals.
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Three patterns, three substitutions
The strategy is pattern recognition.
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Three patterns, three substitutions
core concept
The strategy is pattern recognition. The integrand contains exactly one of three algebraic shapes; each calls for a specific trig substitution because of a specific Pythagorean identity.
Pattern $\sqrt{a^2 - x^2}$ let $x = a\sin\theta$. Then $a^2 - x^2 = a^2(1 - \sin^2\theta) = a^2\cos^2\theta$, so $\sqrt{a^2-x^2} = a\cos\theta$. Also $dx = a\cos\theta\,d\theta$.
Pattern $a^2 + x^2$ let $x = a\tan\theta$. Then $a^2 + x^2 = a^2(1 + \tan^2\theta) = a^2\sec^2\theta$. Also $dx = a\sec^2\theta\,d\theta$.
Pattern $\sqrt{x^2 - a^2}$ let $x = a\sec\theta$. Then $x^2 - a^2 = a^2(\sec^2\theta - 1) = a^2\tan^2\theta$, so $\sqrt{x^2-a^2} = a\tan\theta$. Also $dx = a\sec\theta\tan\theta\,d\theta$.
Worked through the hook: $\displaystyle\int \frac{dx}{\sqrt{9-x^2}}$. Pattern $\sqrt{a^2-x^2}$ with $a=3$. Let $x = 3\sin\theta$, $dx = 3\cos\theta\,d\theta$, $\sqrt{9-x^2} = 3\cos\theta$.
Back-substitute: $\sin\theta = x/3$ so $\theta = \arcsin(x/3)$. Final answer: $\arcsin(x/3) + C$.
(This matches the standard result $\int dx/\sqrt{a^2-x^2} = \arcsin(x/a) + C$.)
Why the radical collapses. Each substitution rewrites the awkward algebraic expression as $a^2 \times (\text{a perfect square of a trig function})$. Square-rooting then gives a single trig function, no radicals left to integrate.
Three patterns table: $\sqrt{a^2-x^2}$ → $\sin$; $a^2+x^2$ → $\tan$; $\sqrt{x^2-a^2}$ → $\sec$ · Each $dx$: $a\cos\theta\,d\theta$ / $a\sec^2\theta\,d\theta$ / $a\sec\theta\tan\theta\,d\theta$ · Pythagorean identity drives the collapse · Always state the principal range of $\theta$ so the square root is positive
Pause, copy the three substitution patterns and their $dx$ substitutions ($a\cos\theta\,d\theta$, $a\sec^2\theta\,d\theta$, $a\sec\theta\tan\theta\,d\theta$), the Pythagorean identity that drives the collapse, and the principal-range requirement into your book.
Quick check: Which substitution best simplifies $\displaystyle\int \frac{dx}{\sqrt{25 - x^2}}$?
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Converting back: the reference triangle
We just saw the three trig-substitution patterns: $\sqrt{a^2-x^2} \to x=a\sin\theta$, $a^2+x^2 \to x=a\tan\theta$, $\sqrt{x^2-a^2} \to x=a\sec\theta$, each collapsing via a Pythagorean identity.
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Converting back: the reference triangle
core concept
We just saw the three trig-substitution patterns: $\sqrt{a^2-x^2} \to x=a\sin\theta$, $a^2+x^2 \to x=a\tan\theta$, $\sqrt{x^2-a^2} \to x=a\sec\theta$, each collapsing via a Pythagorean identity. That raises a question: after integrating in $\theta$, how do we convert the answer back to $x$? This card answers it → draw the reference triangle with sides labelled in terms of $x$ and $a$, then read off every trig ratio needed.
After integrating in $\theta$ the answer often contains $\sin\theta$, $\cos\theta$, $\tan\theta$, but the question was asked in $x$. The fastest route back is to draw the reference triangle implied by the substitution and read off every trig ratio.
Common mistake. Forgetting to convert back. An answer left in terms of $\theta$ is incomplete because $\theta$ is not the original variable. Always finish by replacing $\theta$ and every trig function of $\theta$ with expressions in $x$.
Draw the reference triangle as soon as you substitute · Label the three sides in terms of $x$ and $a$ using the Pythagorean theorem · Read off every trig ratio you need from the triangle · Final answer must be in $x$, replace $\theta$ via $\arcsin$, $\arctan$ or $\text{arcsec}$ as appropriate
Pause, copy the reference-triangle method (draw immediately, label sides using Pythagoras, read trig ratios), and the instruction to express the final answer in $x$ via $\arcsin$, $\arctan$, or $\text{arcsec}$ as appropriate into your book.
Did you get this? True or false: if $x = 4\tan\theta$, then $\sqrt{16 + x^2} = 4\sec\theta$ (taking $\theta \in (-\pi/2, \pi/2)$).
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Worked examples · 3 in a row, reveal as you go
Worked examples · 3 in a row, reveal as you go
Worked examples · 3 in a row, reveal as you go
PROBLEM 1 · SINE SUBSTITUTION
Evaluate $\displaystyle\int \sqrt{4 - x^2}\,dx$ using a trigonometric substitution.
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Pattern $\sqrt{a^2 - x^2}$ with $a = 2$. Let $x = 2\sin\theta$, $\theta \in [-\pi/2, \pi/2]$, so $dx = 2\cos\theta\,d\theta$ and $\sqrt{4-x^2} = 2\cos\theta$.
Identify $a$ from the constant under the radical, then write down both $x$ and $dx$ before substituting.
The $\sec^2\theta$ factor in $dx$ cancels exactly with the $\sec^2\theta$ from the denominator, that is the whole point of this substitution.
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From $x = 3\tan\theta$: $\theta = \arctan(x/3)$. Final: $\displaystyle\frac{1}{3}\arctan\!\frac{x}{3} + C$. This matches the standard form $\int \frac{dx}{a^2+x^2} = \frac{1}{a}\arctan(x/a) + C$.
Memorise this standard result, it derives directly from the tangent substitution and saves time on quizzes.
PROBLEM 3 · SECANT SUBSTITUTION
Evaluate $\displaystyle\int \frac{dx}{x\sqrt{x^2 - 4}}$ for $x \geq 2$, using a trigonometric substitution.
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Pattern $\sqrt{x^2 - a^2}$ with $a = 2$. Let $x = 2\sec\theta$, $\theta \in [0, \pi/2)$, so $dx = 2\sec\theta\tan\theta\,d\theta$ and $\sqrt{x^2-4} = 2\tan\theta$.
The $\sqrt{x^2 - a^2}$ shape is the trademark of the secant substitution. Note $x \geq a$ matches $\sec\theta \geq 1$.
All trig factors cancel, leaving a constant integrand. This is why secant substitution is so clean for this form.
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From $\sec\theta = x/2$: $\theta = \text{arcsec}(x/2) = \arccos(2/x)$. Final: $\displaystyle\frac{1}{2}\arccos\!\frac{2}{x} + C$ (equivalent to $\frac{1}{2}\text{arcsec}(x/2)+C$).
$\text{arcsec}$ is not always listed; convert via $\text{arcsec}(u) = \arccos(1/u)$ for a clean answer in standard inverse-trig form.
Fill the gap: For the integral $\int \frac{dx}{\sqrt{a^2 - x^2}}$ use $x = a\sin\theta$. Then $dx = $ $\theta\,d\theta$ and $\sqrt{a^2-x^2} = $ $\theta$.
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Misconceptions to fix · the 3 traps that cost marks
Misconceptions to fix · the 3 traps that cost marks
Misconceptions to fix · the 3 traps that cost marks
Trap 01
Forgetting to substitute $dx$
Beginners often replace only $x$ and the radical, then keep the original $dx$. Every change of variable must include $dx = a\cos\theta\,d\theta$ (or the tan/sec version). Missing this kills the cancellation and leads to a nonsense integral.
Trap 02
Wrong sign on the radical
$\sqrt{a^2\cos^2\theta} = a|\cos\theta|$, not $a\cos\theta$, in general. The principal range ($\theta \in [-\pi/2, \pi/2]$ for sine, $[0,\pi/2)$ for secant) is chosen exactly so $\cos\theta \geq 0$ (or $\tan\theta \geq 0$). State the range to justify dropping the absolute value.
Trap 03
Leaving the answer in $\theta$
The original variable is $x$. An answer like $2\theta + \sin 2\theta + C$ is incomplete. Use the reference triangle to express every trig function of $\theta$ in terms of $x$, and write $\theta$ itself as $\arcsin(x/a)$, $\arctan(x/a)$ or $\text{arcsec}(x/a)$.
Did you get this? True or false: for the integral $\int dx/(x^2 + 16)$ the appropriate substitution is $x = 4\sin\theta$.
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Revisit your thinking
Activities · practice with the ideas
Activities · practice with the ideas
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Evaluate $\displaystyle\int \frac{dx}{\sqrt{1 - x^2}}$ using a trig substitution. Confirm you obtain a familiar inverse trig function.
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Evaluate $\displaystyle\int \frac{dx}{4 + x^2}$ via the tangent substitution. Compare with the standard form.
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Evaluate $\displaystyle\int \frac{x^2}{\sqrt{1 - x^2}}\,dx$ using $x = \sin\theta$, and finish with a double-angle identity.
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Using $x = a\sec\theta$, derive the result $\displaystyle\int \frac{dx}{x\sqrt{x^2 - a^2}} = \frac{1}{a}\text{arcsec}\!\frac{x}{a} + C$.
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Match each radical to the correct substitution and write the resulting expression: (a) $\sqrt{36 - x^2}$, (b) $\sqrt{x^2 - 1}$, (c) $\sqrt{x^2 + 100}$.
Odd one out: Three of the following pairs (radical, substitution) are correctly matched. Which one is NOT?
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Revisit your thinking
Earlier you tried $\displaystyle\int \frac{dx}{\sqrt{9 - x^2}}$ and noticed $u$-substitution had no inner derivative to seize.
With $x = 3\sin\theta$, the radical collapses to $3\cos\theta$, the $dx = 3\cos\theta\,d\theta$ cancels it exactly, and the integral reduces to $\int d\theta = \theta + C = \arcsin(x/3) + C$. The deeper lesson is that the choice of substitution is dictated by the algebraic shape of the integrand, not by guesswork. Three radical shapes → three substitutions → three Pythagorean identities. Internalise the table.
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Show what you have learned
Multiple choice, then short answer under exam conditions.
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Multiple choice
+5 XP per correct · +25 XP all-correct
Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.
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Short answer
ApplyBand 32 marks
Q1. Use a trigonometric substitution to evaluate $\displaystyle\int \frac{dx}{\sqrt{16 - x^2}}$. (2 marks)
Q3. Show that $\displaystyle\int \sqrt{9 - x^2}\,dx = \frac{9}{2}\arcsin\!\frac{x}{3} + \frac{x\sqrt{9-x^2}}{2} + C$ using a trigonometric substitution. (3 marks)
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Comprehensive answers (click to reveal)
Activity answers:
1. $x = \sin\theta$: integrand becomes $\int d\theta = \theta + C = \arcsin x + C$.
