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Module 8 · L4 of 20 ~40 min ⚡ +95 XP available

Trigonometric Substitutions

The expression $\sqrt{a^2 - x^2}$ cannot be handled by a plain $u$-substitution — but if you substitute $x = a\sin\theta$, the radical vanishes via the Pythagorean identity $1 - \sin^2\theta = \cos^2\theta$. This lesson teaches you the two key trig substitutions and exactly when to deploy each one.

Today's hook — Why can't you use a plain $u$-substitution on $\displaystyle\int \sqrt{4 - x^2}\,dx$? Write your reasoning before reading on. You'll revisit this on card 11.

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Recall — your gut answer first

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Consider $\displaystyle\int \sqrt{4 - x^2}\,dx$. Without calculating — why won't a standard $u = 4 - x^2$ substitution complete this integral cleanly? Write your reasoning.

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The two trig substitutions

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Two Pythagorean identities drive the substitutions. Match the radical form to the substitution:

Form 1: $\sqrt{a^2 - x^2}$ → set $x = a\sin\theta$

$a^2 - x^2 = a^2 - a^2\sin^2\theta = a^2\cos^2\theta$

Form 2: $\sqrt{a^2 + x^2}$ → set $x = a\tan\theta$

$a^2 + x^2 = a^2 + a^2\tan^2\theta = a^2\sec^2\theta$

$dx = a\cos\theta\,d\theta$ or $dx = a\sec^2\theta\,d\theta$

Minus sign → sine

$a^2 - x^2$ means $\sin^2\theta + \cos^2\theta = 1$ in disguise. Use $x = a\sin\theta$.

Plus sign → tangent

$a^2 + x^2$ means $1 + \tan^2\theta = \sec^2\theta$ in disguise. Use $x = a\tan\theta$.

Back-substitution

After integrating in $\theta$, draw a right triangle to express $\sin\theta$, $\cos\theta$, $\tan\theta$ back in terms of $x$.

What you'll master

Know

Key facts

$\sqrt{a^2-x^2}$ requires $x = a\sin\theta$; $\sqrt{a^2+x^2}$ requires $x = a\tan\theta$
$dx = a\cos\theta\,d\theta$ (sine sub) and $dx = a\sec^2\theta\,d\theta$ (tangent sub)
The Pythagorean identities $\sin^2\theta + \cos^2\theta = 1$ and $1 + \tan^2\theta = \sec^2\theta$ underpin both substitutions

Understand

Concepts

Why a trig substitution eliminates the radical — the identity does the work
How a right triangle connects $\theta$ back to $x$ for back-substitution
When each form applies, keyed to the $\pm$ sign under the radical

Can do

Skills

Apply $x = a\sin\theta$ or $x = a\tan\theta$ to eliminate a radical
Complete the integral in $\theta$ and back-substitute using a triangle
Convert limits for definite integrals using the inverse trig function

Key terms

Trigonometric substitutionA substitution of the form $x = a\sin\theta$ or $x = a\tan\theta$ used to eliminate a square-root expression involving $a^2 \pm x^2$.

$x = a\sin\theta$Used when the integrand contains $\sqrt{a^2-x^2}$. Gives $dx = a\cos\theta\,d\theta$ and $\sqrt{a^2-x^2} = a\cos\theta$.

$x = a\tan\theta$Used when the integrand contains $\sqrt{a^2+x^2}$. Gives $dx = a\sec^2\theta\,d\theta$ and $\sqrt{a^2+x^2} = a\sec\theta$.

Pythagorean identities$\sin^2\theta + \cos^2\theta = 1$ and $1 + \tan^2\theta = \sec^2\theta$. These are used to simplify $a^2 - x^2$ and $a^2 + x^2$ after substitution.

Reference triangleA right triangle labelled with sides $x$, $a$, and $\sqrt{a^2 \pm x^2}$ to convert trig functions of $\theta$ back into expressions in $x$.

Domain restrictionFor $x = a\sin\theta$, restrict $\theta \in [-\pi/2, \pi/2]$; for $x = a\tan\theta$, restrict $\theta \in (-\pi/2, \pi/2)$ to ensure the substitution is one-to-one.

The sine substitution: $\sqrt{a^2 - x^2}$

core concept

When the integrand contains $\sqrt{a^2 - x^2}$, set $x = a\sin\theta$ (with $\theta \in [-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$). Then:

$$a^2 - x^2 = a^2 - a^2\sin^2\theta = a^2(1 - \sin^2\theta) = a^2\cos^2\theta$$

So $\sqrt{a^2 - x^2} = a\cos\theta$ (positive since $\theta \in [-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$), and $dx = a\cos\theta\,d\theta$.

Worked application: $\displaystyle\int \sqrt{4 - x^2}\,dx$ — this answers your hook!

Set $x = 2\sin\theta$, $dx = 2\cos\theta\,d\theta$. Then $\sqrt{4-x^2} = 2\cos\theta$:

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

Back-substitute: $\theta = \sin^{-1}\!\left(\tfrac{x}{2}\right)$ and $\sin 2\theta = 2\sin\theta\cos\theta = 2 \cdot \tfrac{x}{2} \cdot \tfrac{\sqrt{4-x^2}}{2} = \tfrac{x\sqrt{4-x^2}}{2}$.

Final answer: $\displaystyle 2\sin^{-1}\!\left(\frac{x}{2}\right) + \frac{x\sqrt{4-x^2}}{2} + C$

Why plain $u$-sub fails here. Setting $u = 4 - x^2$ gives $du = -2x\,dx$, leaving $\sqrt{u}$ divided by $\sqrt{u}$ — which does not simplify without an extra factor. The trig substitution is essential because the radical has no derivative sitting alongside it.

When the integrand contains $\sqrt{a^2 - x^2}$, set $x = a\sin\theta$ (with $\theta \in [-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$). Then:

Pause — copy the sine substitution: $x=a\sin\theta$, $dx=a\cos\theta\,d\theta$, $\sqrt{a^2-x^2}=a\cos\theta$, with a worked integral into your book.

Quick check: For $\displaystyle\int \frac{1}{\sqrt{9 - x^2}}\,dx$, the correct substitution is:

The tangent substitution: $\sqrt{a^2 + x^2}$

core concept

We just saw the sine substitution: set $x=a\sin\theta$, so $dx=a\cos\theta\,d\theta$ and $\sqrt{a^2-x^2}=a\cos\theta$, converting the radical to a trig expression. That raises a question: for $\sqrt{a^2+x^2}$, the sine substitution fails — which identity removes this radical, and what substitution gives $dx$ and $\sqrt{a^2+x^2}$ in terms of $\theta$? This card answers it → set $x=a\tan\theta$; then $\sqrt{a^2+x^2}=a\sec\theta$ and $dx=a\sec^2\theta\,d\theta$.

When the integrand contains $\sqrt{a^2 + x^2}$ (or just $a^2 + x^2$ with no radical), set $x = a\tan\theta$ (with $\theta \in (-\tfrac{\pi}{2}, \tfrac{\pi}{2})$). Then:

$$a^2 + x^2 = a^2 + a^2\tan^2\theta = a^2(1 + \tan^2\theta) = a^2\sec^2\theta$$

So $\sqrt{a^2+x^2} = a\sec\theta$ (positive since $\theta \in (-\tfrac{\pi}{2}, \tfrac{\pi}{2})$), and $dx = a\sec^2\theta\,d\theta$.

Example: $\displaystyle\int \frac{1}{a^2 + x^2}\,dx$

Set $x = a\tan\theta$, $dx = a\sec^2\theta\,d\theta$:

$$\int \frac{a\sec^2\theta}{a^2\sec^2\theta}\,d\theta = \frac{1}{a}\int 1\,d\theta = \frac{\theta}{a} + C = \frac{1}{a}\tan^{-1}\!\left(\frac{x}{a}\right) + C$$

This is a standard result you will use repeatedly in later lessons.

The tangent sub doesn't only apply to square roots. The integrand $\dfrac{1}{a^2+x^2}$ has no radical, yet the substitution $x = a\tan\theta$ perfectly eliminates the denominator. Recognise the pattern $a^2 + x^2$ regardless of whether a radical appears.

When the integrand contains $\sqrt{a^2 + x^2}$ (or just $a^2 + x^2$ with no radical), set $x = a\tan\theta$ (with $\theta \in (-\tfrac{\pi}{2}, \tfrac{\pi}{2})$). Then:

Pause — copy the tangent substitution: $x=a\tan\theta$, $dx=a\sec^2\theta\,d\theta$, $\sqrt{a^2+x^2}=a\sec\theta$, with a worked integral into your book.

Did you get this? True or false: after substituting $x = a\tan\theta$, the expression $a^2 + x^2$ simplifies to $a^2\sec^2\theta$.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · SINE SUBSTITUTION

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

$\sqrt{9-x^2}$ has form $\sqrt{a^2-x^2}$ with $a=3$. Set $x = 3\sin\theta$, $dx = 3\cos\theta\,d\theta$.

Match the radical to the sine substitution form. Identify $a = 3$ so that $a^2 = 9$.

PROBLEM 2 · TANGENT SUBSTITUTION

Find $\displaystyle\int \frac{1}{\sqrt{x^2+16}}\,dx$.

$\sqrt{x^2+16} = \sqrt{x^2+4^2}$: form $\sqrt{a^2+x^2}$ with $a=4$. Set $x = 4\tan\theta$, $dx = 4\sec^2\theta\,d\theta$.

Note: $x^2 + 4^2$ not $4^2 + x^2$ — order in the sum doesn't matter; identify $a^2 = 16$.

PROBLEM 3 · DEFINITE INTEGRAL

Evaluate $\displaystyle\int_0^{2} \frac{x^2}{\sqrt{4-x^2}}\,dx$.

Set $x = 2\sin\theta$, $dx = 2\cos\theta\,d\theta$. Limits: $x=0\Rightarrow\theta=0$; $x=2\Rightarrow\theta=\tfrac{\pi}{2}$.

Convert limits via $\theta = \sin^{-1}(x/2)$: $\sin^{-1}(0)=0$, $\sin^{-1}(1)=\tfrac{\pi}{2}$.

Fill the gap: After substituting $x = 5\sin\theta$ into $\sqrt{25 - x^2}$, the radical simplifies to $5\cos\theta$ using the identity $1 - \sin^2\theta = $ .

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Using the wrong substitution

$\sqrt{a^2 - x^2}$ needs $x = a\sin\theta$; $\sqrt{a^2 + x^2}$ needs $x = a\tan\theta$. Mixing them means the Pythagorean identity doesn't apply and the radical won't simplify. Match the sign first.

Trap 02

Forgetting to substitute $dx$

With $x = a\sin\theta$ you must also replace $dx$ with $a\cos\theta\,d\theta$. Leaving $dx$ in the integral produces a mixed-variable expression that cannot be evaluated.

Trap 03

Back-substitution errors

After integrating in $\theta$, you must convert back to $x$. Draw a reference triangle labelled with $x$, $a$, and $\sqrt{a^2 \pm x^2}$, then read off $\sin\theta$, $\cos\theta$, $\tan\theta$ directly. Don't try to do this from memory.

Did you get this? True or false: the substitution $x = 2\tan\theta$ is appropriate for $\displaystyle\int \frac{1}{\sqrt{4 + x^2}}\,dx$.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

For each integral, state which substitution ($x = a\sin\theta$ or $x = a\tan\theta$) you would use and identify the value of $a$: (i) $\int\sqrt{25-x^2}\,dx$ (ii) $\int\frac{1}{9+x^2}\,dx$

Find $\displaystyle\int \frac{1}{\sqrt{16 - x^2}}\,dx$. State the substitution, simplify, integrate and back-substitute.

Find $\displaystyle\int \frac{1}{4 + x^2}\,dx$ using $x = 2\tan\theta$.

Evaluate $\displaystyle\int_0^{3} \frac{1}{\sqrt{9-x^2}}\,dx$. Convert the limits using $\theta = \sin^{-1}(x/3)$.

Draw a labelled reference triangle for $x = 5\sin\theta$ and write expressions for $\cos\theta$ and $\tan\theta$ in terms of $x$.

Odd one out: Three of these pairs (integral, substitution) are correctly matched. Which one is NOT?

Revisit your thinking

Earlier you considered why plain $u$-substitution fails for $\displaystyle\int \sqrt{4-x^2}\,dx$.

Setting $u = 4-x^2$ gives $du = -2x\,dx$, so $dx = -du/(2x) = -du/(2\sqrt{4-u})$ — the $x$ doesn't cancel, leaving a harder integral. The trig substitution $x = 2\sin\theta$ works because $1 - \sin^2\theta = \cos^2\theta$ makes the radical vanish exactly. Did your hook answer identify this?

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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.

Short answer

ApplyBand 31 mark

Q1. State the substitution and $dx$ expression for integrating $\displaystyle\int \frac{1}{\sqrt{25-x^2}}\,dx$. (1 mark)

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ApplyBand 42 marks

Q2. Find $\displaystyle\int \frac{1}{x^2+9}\,dx$ using $x = 3\tan\theta$. Show full working. (2 marks)

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AnalyseBand 53 marks

Q3. Evaluate $\displaystyle\int_0^{1} \frac{1}{\sqrt{1-x^2}}\,dx$. Change the limits and state the geometric meaning of your answer. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. (i) $x = 5\sin\theta$, $a=5$; (ii) $x = 3\tan\theta$, $a=3$. 2. $x=4\sin\theta$, $dx=4\cos\theta\,d\theta$, $\sqrt{16-x^2}=4\cos\theta$; integral $= \int 1\,d\theta = \theta + C = \sin^{-1}(x/4)+C$. 3. $x=2\tan\theta$, $dx=2\sec^2\theta\,d\theta$; $4+x^2=4\sec^2\theta$; $\frac{2\sec^2\theta\,d\theta}{4\sec^2\theta} = \frac{1}{2}\theta + C = \frac{1}{2}\tan^{-1}(x/2)+C$. 4. Limits: $\theta=0$ to $\theta=\pi/2$; $[\theta]_0^{\pi/2} = \pi/2$. 5. Adj $= \sqrt{25-x^2}$; $\cos\theta = \sqrt{25-x^2}/5$; $\tan\theta = x/\sqrt{25-x^2}$.

Q1 (1 mark): $x = 5\sin\theta$, $dx = 5\cos\theta\,d\theta$; gives $\sin^{-1}(x/5)+C$ [1].

Q2 (2 marks): $x=3\tan\theta$, $dx=3\sec^2\theta\,d\theta$, $x^2+9=9\sec^2\theta$ [1]; integral $= \frac{1}{3}\int 1\,d\theta = \frac{\theta}{3}+C = \frac{1}{3}\tan^{-1}(x/3)+C$ [1].

Q3 (3 marks): $x=\sin\theta$; limits $\theta=0$ to $\theta=\pi/2$ [1]; $\int_0^{\pi/2}1\,d\theta = \pi/2$ [1]; geometric meaning: this is the area under $y=1/\sqrt{1-x^2}$ from 0 to 1, equivalent to a quarter of the unit circle's arc — the answer $\pi/2$ equals $\sin^{-1}(1)-\sin^{-1}(0)$ [1].

Boss battle · The Trig Sub Terminator

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Science Jump · platform challenge

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Module overview · Extension 1 · Checkpoint 1