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Module 15 · L07 of 16 ~40 min ⚡ +90 XP available

Trigonometric Substitutions I

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

Recall — your gut answer first

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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

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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.

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

$\sqrt{a^2-x^2} \;\xrightarrow{\,x = a\sin\theta\,}\; a\cos\theta$

$\sqrt{a^2-x^2}$ → sin

Set $x = a\sin\theta$, $-\pi/2 \leq \theta \leq \pi/2$. Then $\sqrt{a^2-x^2} = a\cos\theta$ (positive on this range) and $dx = a\cos\theta\,d\theta$.

$a^2+x^2$ → tan

Set $x = a\tan\theta$, $-\pi/2 < \theta < \pi/2$. Then $a^2+x^2 = a^2\sec^2\theta$ and $dx = a\sec^2\theta\,d\theta$.

$\sqrt{x^2-a^2}$ → sec

Set $x = a\sec\theta$, $0 \leq \theta < \pi/2$ (for $x \geq a$). Then $\sqrt{x^2-a^2} = a\tan\theta$ and $dx = a\sec\theta\tan\theta\,d\theta$.

What you'll master

Know

Key facts

$\sin^2\theta + \cos^2\theta = 1$, $1+\tan^2\theta = \sec^2\theta$
The three radical patterns and their matching substitutions
$dx$ formulas: $a\cos\theta\,d\theta$, $a\sec^2\theta\,d\theta$, $a\sec\theta\tan\theta\,d\theta$
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

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.

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 $x = 3\sin\theta$, $dx = 3\cos\theta\,d\theta$, $\sqrt{9-x^2} = 3\cos\theta$.

$\displaystyle\int \frac{3\cos\theta}{3\cos\theta}\,d\theta = \int 1\,d\theta = \theta + C$.
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 $\displaystyle\int \frac{dx}{\sqrt{25 - x^2}}$?

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.

If $x = a\sin\theta$: opposite $= x$, hypotenuse $= a$, adjacent $= \sqrt{a^2 - x^2}$. Then $\cos\theta = \frac{\sqrt{a^2-x^2}}{a}$, $\tan\theta = \frac{x}{\sqrt{a^2-x^2}}$.
If $x = a\tan\theta$: opposite $= x$, adjacent $= a$, hypotenuse $= \sqrt{a^2 + x^2}$. Then $\sin\theta = \frac{x}{\sqrt{a^2+x^2}}$, $\sec\theta = \frac{\sqrt{a^2+x^2}}{a}$.
If $x = a\sec\theta$: hypotenuse $= x$, adjacent $= a$, opposite $= \sqrt{x^2 - a^2}$. Then $\tan\theta = \frac{\sqrt{x^2-a^2}}{a}$, $\sin\theta = \frac{\sqrt{x^2-a^2}}{x}$.

$$x = a\sin\theta \;\Longrightarrow\; \sin\theta = \tfrac{x}{a},\quad \cos\theta = \tfrac{\sqrt{a^2-x^2}}{a},\quad \tan\theta = \tfrac{x}{\sqrt{a^2-x^2}}$$

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 $\theta \in (-\pi/2, \pi/2)$).

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.

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.

PROBLEM 2 · TANGENT SUBSTITUTION

Evaluate $\displaystyle\int \frac{dx}{9 + x^2}$ using a trigonometric substitution. Verify it matches the standard form.

Pattern $a^2 + x^2$ with $a = 3$. Let $x = 3\tan\theta$, $\theta \in (-\pi/2, \pi/2)$, so $dx = 3\sec^2\theta\,d\theta$ and $9 + x^2 = 9(1 + \tan^2\theta) = 9\sec^2\theta$.

Even without a radical, $a^2+x^2$ is the tangent-substitution pattern because $1+\tan^2\theta = \sec^2\theta$.

PROBLEM 3 · SECANT SUBSTITUTION

Evaluate $\displaystyle\int \frac{dx}{x\sqrt{x^2 - 4}}$ for $x \geq 2$, using a trigonometric substitution.

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 $\sec\theta \geq 1$.

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} = $ $\theta$.

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 $x = 4\sin\theta$.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Evaluate $\displaystyle\int \frac{dx}{\sqrt{1 - x^2}}$ using a trig substitution. Confirm you obtain a familiar inverse trig function.

Evaluate $\displaystyle\int \frac{dx}{4 + x^2}$ via the tangent substitution. Compare with the standard form.

Evaluate $\displaystyle\int \frac{x^2}{\sqrt{1 - x^2}}\,dx$ using $x = \sin\theta$, and finish with a double-angle identity.

Using $x = a\sec\theta$, derive the result $\displaystyle\int \frac{dx}{x\sqrt{x^2 - a^2}} = \frac{1}{a}\text{arcsec}\!\frac{x}{a} + C$.

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?

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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Multiple choice

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Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 32 marks

Q1. Use a trigonometric substitution to evaluate $\displaystyle\int \frac{dx}{\sqrt{16 - x^2}}$. (2 marks)

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ApplyBand 43 marks

Q2. Evaluate $\displaystyle\int \frac{dx}{(x^2 + 1)^{3/2}}$ using $x = \tan\theta$. (3 marks)

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AnalyseBand 53 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 $\int d\theta = \theta + C = \arcsin x + C$.

2. $x = 2\tan\theta$: integrand becomes $\int (2\sec^2\theta)/(4\sec^2\theta)\,d\theta = \tfrac{1}{2}\theta + C = \tfrac{1}{2}\arctan(x/2) + C$.

3. $x = \sin\theta$: integrand becomes $\int \sin^2\theta\,d\theta = \tfrac{1}{2}\theta - \tfrac{1}{4}\sin 2\theta + C = \tfrac{1}{2}\arcsin x - \tfrac{1}{2}x\sqrt{1-x^2} + C$.

4. $x = a\sec\theta$: integrand becomes $\int (1/a)\,d\theta = \theta/a + C = (1/a)\text{arcsec}(x/a) + C$.

5. (a) $x = 6\sin\theta$, $\sqrt{36-x^2} = 6\cos\theta$; (b) $x = \sec\theta$, $\sqrt{x^2-1} = \tan\theta$; (c) $x = 10\tan\theta$, $\sqrt{x^2+100} = 10\sec\theta$.

Q1 (2 marks): Let $x = 4\sin\theta$; $dx = 4\cos\theta\,d\theta$; $\sqrt{16-x^2} = 4\cos\theta$ [1]. Integral $= \int d\theta = \arcsin(x/4) + C$ [1].

Q2 (3 marks): $x = \tan\theta$, $dx = \sec^2\theta\,d\theta$, $(x^2+1)^{3/2} = \sec^3\theta$ [1]. Integrand $= \int \cos\theta\,d\theta = \sin\theta + C$ [1]. Reference triangle gives $\sin\theta = x/\sqrt{1+x^2}$, so answer $= x/\sqrt{1+x^2} + C$ [1].

Q3 (3 marks): $x = 3\sin\theta$ gives $\sqrt{9-x^2} = 3\cos\theta$ and $dx = 3\cos\theta\,d\theta$; integrand becomes $9\cos^2\theta\,d\theta$ [1]. Using $\cos^2\theta = \tfrac{1}{2}(1+\cos 2\theta)$: $\int = \tfrac{9}{2}\theta + \tfrac{9}{4}\sin 2\theta + C$ [1]. Back-substitute $\theta = \arcsin(x/3)$ and $\sin 2\theta = 2 \cdot (x/3) \cdot (\sqrt{9-x^2}/3) = 2x\sqrt{9-x^2}/9$ to give $\tfrac{9}{2}\arcsin(x/3) + x\sqrt{9-x^2}/2 + C$ [1].

Boss battle · The Substitution Sentinel

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Five timed questions on choosing and executing trig substitutions for $\sqrt{a^2-x^2}$, $a^2+x^2$ and $\sqrt{x^2-a^2}$. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

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Science Jump · platform challenge

Climb platforms by answering quick trig-substitution questions. Lighter alternative to the boss.

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Module overview · Extension 2