Trigonometric Substitution

Using trig identities to simplify difficult integrals

🔺 Trigonometric Substitution

The Problem

How do we integrate expressions with square roots like $\sqrt{a^2 - x^2}$ , $\sqrt{a^2 + x^2}$ , or $\sqrt{x^2 - a^2}$ ?

Example: $\int \frac{1}{\sqrt{4-x^2}}\,dx$

💡 Key Idea: Use trig substitutions to eliminate the square root using Pythagorean identities!

The Three Cases

Case 1: $\sqrt{a^2 - x^2}$

Substitution: $x = a\sin\theta$

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

Then: $\sqrt{a^2 - x^2} = a\cos\theta$ (assuming $\cos\theta > 0$ )

Also: $dx = a\cos\theta\,d\theta$

Case 2: $\sqrt{a^2 + x^2}$

Substitution: $x = a\tan\theta$

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

Then: $\sqrt{a^2 + x^2} = a\sec\theta$ (assuming $\sec\theta > 0$ )

Also: $dx = a\sec^2\theta\,d\theta$

Case 3: $\sqrt{x^2 - a^2}$

Substitution: $x = a\sec\theta$

Why: $x^2 - a^2 = a^2\sec^2\theta - a^2 = a^2(\sec^2\theta - 1) = a^2\tan^2\theta$

Then: $\sqrt{x^2 - a^2} = a\tan\theta$ (assuming $\tan\theta > 0$ )

Also: $dx = a\sec\theta\tan\theta\,d\theta$

Quick Reference Table

| Expression | Substitution | Identity Used | Result | |------------|-------------|---------------|--------| | $\sqrt{a^2 - x^2}$ | $x = a\sin\theta$ | $\cos^2\theta = 1 - \sin^2\theta$ | $a\cos\theta$ | | $\sqrt{a^2 + x^2}$ | $x = a\tan\theta$ | $\sec^2\theta = 1 + \tan^2\theta$ | $a\sec\theta$ | | $\sqrt{x^2 - a^2}$ | $x = a\sec\theta$ | $\tan^2\theta = \sec^2\theta - 1$ | $a\tan\theta$ |

Example 1: Case 1 - $\sqrt{a^2 - x^2}$

Evaluate $\int \frac{1}{\sqrt{4-x^2}}\,dx$

Step 1: Identify the form

$\sqrt{4-x^2} = \sqrt{2^2 - x^2}$ , so $a = 2$

This is Case 1!

Step 2: Make substitution

Let $x = 2\sin\theta$

Then $dx = 2\cos\theta\,d\theta$

And $\sqrt{4-x^2} = \sqrt{4 - 4\sin^2\theta} = 2\sqrt{1-\sin^2\theta} = 2\cos\theta$

Step 3: Rewrite integral

$\int \frac{1}{\sqrt{4-x^2}}\,dx = \int \frac{1}{2\cos\theta} \cdot 2\cos\theta\,d\theta$

$= \int 1\,d\theta = \theta + C$

Step 4: Convert back to x

From $x = 2\sin\theta$ , we get $\sin\theta = \frac{x}{2}$

Therefore $\theta = \arcsin\left(\frac{x}{2}\right)$

Answer: $\arcsin\left(\frac{x}{2}\right) + C$

Example 2: Case 2 - $\sqrt{a^2 + x^2}$

Evaluate $\int \frac{x^2}{\sqrt{1+x^2}}\,dx$

Step 1: Identify the form

$\sqrt{1+x^2} = \sqrt{1^2 + x^2}$ , so $a = 1$

This is Case 2!

Step 2: Make substitution

Let $x = \tan\theta$

Then $dx = \sec^2\theta\,d\theta$

And $\sqrt{1+x^2} = \sqrt{1 + \tan^2\theta} = \sec\theta$

Step 3: Rewrite integral

$\int \frac{x^2}{\sqrt{1+x^2}}\,dx = \int \frac{\tan^2\theta}{\sec\theta} \cdot \sec^2\theta\,d\theta$

$= \int \tan^2\theta \cdot \sec\theta\,d\theta$

Step 4: Use identity $\tan^2\theta = \sec^2\theta - 1$

$= \int (\sec^2\theta - 1)\sec\theta\,d\theta$

$= \int (\sec^3\theta - \sec\theta)\,d\theta$

Step 5: Integrate

These are standard integrals (using integration by parts for $\sec^3\theta$ ):

$\int \sec^3\theta\,d\theta = \frac{1}{2}(\sec\theta\tan\theta + \ln|\sec\theta + \tan\theta|) + C$

$\int \sec\theta\,d\theta = \ln|\sec\theta + \tan\theta| + C$

$= \frac{1}{2}\sec\theta\tan\theta + \frac{1}{2}\ln|\sec\theta + \tan\theta| - \ln|\sec\theta + \tan\theta| + C$

$= \frac{1}{2}\sec\theta\tan\theta - \frac{1}{2}\ln|\sec\theta + \tan\theta| + C$

Step 6: Convert back to x

From $x = \tan\theta$ :

$\tan\theta = x$
$\sec\theta = \sqrt{1+x^2}$

$= \frac{1}{2}\sqrt{1+x^2} \cdot x - \frac{1}{2}\ln|\sqrt{1+x^2} + x| + C$

Answer: $\frac{x\sqrt{1+x^2}}{2} - \frac{1}{2}\ln|\sqrt{1+x^2} + x| + C$

Example 3: Case 3 - $\sqrt{x^2 - a^2}$

Evaluate $\int \sqrt{x^2 - 9}\,dx$

Step 1: Identify the form

$\sqrt{x^2 - 9} = \sqrt{x^2 - 3^2}$ , so $a = 3$

This is Case 3!

Step 2: Make substitution

Let $x = 3\sec\theta$

Then $dx = 3\sec\theta\tan\theta\,d\theta$

And $\sqrt{x^2-9} = \sqrt{9\sec^2\theta - 9} = 3\sqrt{\sec^2\theta - 1} = 3\tan\theta$

Step 3: Rewrite integral

$\int \sqrt{x^2-9}\,dx = \int 3\tan\theta \cdot 3\sec\theta\tan\theta\,d\theta$

$= 9\int \tan^2\theta\sec\theta\,d\theta$

This is the same type as Example 2!

$= 9\int (\sec^2\theta - 1)\sec\theta\,d\theta$

$= 9\left[\frac{1}{2}\sec\theta\tan\theta - \frac{1}{2}\ln|\sec\theta + \tan\theta|\right] + C$

Step 4: Convert back to x

From $x = 3\sec\theta$ :

$\sec\theta = \frac{x}{3}$
$\tan\theta = \sqrt{\sec^2\theta - 1} = \sqrt{\frac{x^2}{9} - 1} = \frac{\sqrt{x^2-9}}{3}$

$= \frac{9}{2} \cdot \frac{x}{3} \cdot \frac{\sqrt{x^2-9}}{3} - \frac{9}{2}\ln\left|\frac{x}{3} + \frac{\sqrt{x^2-9}}{3}\right| + C$

$= \frac{x\sqrt{x^2-9}}{2} - \frac{9}{2}\ln\left|\frac{x + \sqrt{x^2-9}}{3}\right| + C$

Answer: $\frac{x\sqrt{x^2-9}}{2} - \frac{9}{2}\ln|x + \sqrt{x^2-9}| + C'$ (absorbing constants into C)

Drawing the Reference Triangle

To convert back from $\theta$ to $x$ , draw a reference triangle!

For $x = a\sin\theta$ :

Triangle with:

Opposite = $x$
Hypotenuse = $a$
Adjacent = $\sqrt{a^2 - x^2}$

Then: $\cos\theta = \frac{\sqrt{a^2-x^2}}{a}$ , $\tan\theta = \frac{x}{\sqrt{a^2-x^2}}$

For $x = a\tan\theta$ :

Triangle with:

Opposite = $x$
Adjacent = $a$
Hypotenuse = $\sqrt{a^2 + x^2}$

Then: $\sin\theta = \frac{x}{\sqrt{a^2+x^2}}$ , $\sec\theta = \frac{\sqrt{a^2+x^2}}{a}$

For $x = a\sec\theta$ :

Triangle with:

Hypotenuse = $x$
Adjacent = $a$
Opposite = $\sqrt{x^2 - a^2}$

Then: $\tan\theta = \frac{\sqrt{x^2-a^2}}{a}$ , $\sin\theta = \frac{\sqrt{x^2-a^2}}{x}$

When to Use Trig Substitution

Use when you see:

$\sqrt{a^2 - x^2}$ or $(a^2 - x^2)^{n/2}$
$\sqrt{a^2 + x^2}$ or $(a^2 + x^2)^{n/2}$
$\sqrt{x^2 - a^2}$ or $(x^2 - a^2)^{n/2}$

Don't use when:

Regular u-substitution works
The expression can be simplified algebraically first
Integration by parts is simpler

⚠️ Common Mistakes

Mistake 1: Wrong Substitution

$\sqrt{4 + x^2}$ → Use $x = 2\tan\theta$ , NOT $x = 2\sin\theta$ !

Match the pattern carefully!

Mistake 2: Forgetting to Convert Back

Your final answer must be in terms of x, not $\theta$ !

Use the reference triangle or inverse trig functions.

Mistake 3: Sign Errors

When taking square roots: $\sqrt{\cos^2\theta} = |\cos\theta|$

Usually we assume the appropriate range for $\theta$ where the trig function is positive.

Mistake 4: Forgetting dx

When substituting $x = a\sin\theta$ , don't forget: $dx = a\cos\theta\,d\theta$

Completing the Square First

Sometimes you need to complete the square before using trig substitution!

Example: $\int \frac{1}{\sqrt{2x - x^2}}\,dx$

Complete the square: $2x - x^2 = -(x^2 - 2x) = -(x^2 - 2x + 1 - 1) = -(x-1)^2 + 1 = 1 - (x-1)^2$

Now it's in the form $a^2 - u^2$ with $u = x-1$ and $a = 1$ .

Standard Trig Integrals to Know

$\int \sec\theta\,d\theta = \ln|\sec\theta + \tan\theta| + C$

$\int \sec^2\theta\,d\theta = \tan\theta + C$

$\int \sec^3\theta\,d\theta = \frac{1}{2}(\sec\theta\tan\theta + \ln|\sec\theta + \tan\theta|) + C$

$\int \tan\theta\,d\theta = -\ln|\cos\theta| + C = \ln|\sec\theta| + C$

📝 Practice Strategy

Identify the pattern: $a^2 - x^2$ , $a^2 + x^2$ , or $x^2 - a^2$
Complete the square if necessary
Choose substitution from the table
Calculate dx (derivative of substitution)
Simplify the square root using Pythagorean identity
Integrate with respect to $\theta$
Draw reference triangle to convert back
Express answer in terms of x

📚 Practice Problems

1Problem 1hard

❓ Question:

Evaluate $\int \frac{x^2}{\sqrt{9-x^2}}\,dx$ using trigonometric substitution.

💡 Show Solution

Step 1: Identify the form

$\sqrt{9-x^2} = \sqrt{3^2 - x^2}$ → Case 1, use $x = 3\sin\theta$

Step 2: Make substitution

$x = 3\sin\theta$ , $dx = 3\cos\theta\,d\theta$

$\sqrt{9-x^2} = 3\cos\theta$

$x^2 = 9\sin^2\theta$

Step 3: Rewrite integral

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

$= \int 9\sin^2\theta\,d\theta$

Step 4: Use identity $\sin^2\theta = \frac{1-\cos 2\theta}{2}$

$= 9\int \frac{1-\cos 2\theta}{2}\,d\theta$

$= \frac{9}{2}\int (1-\cos 2\theta)\,d\theta$

$= \frac{9}{2}\left(\theta - \frac{\sin 2\theta}{2}\right) + C$

$= \frac{9\theta}{2} - \frac{9\sin 2\theta}{4} + C$

Step 5: Use $\sin 2\theta = 2\sin\theta\cos\theta$

$= \frac{9\theta}{2} - \frac{9(2\sin\theta\cos\theta)}{4} + C$

$= \frac{9\theta}{2} - \frac{9\sin\theta\cos\theta}{2} + C$

Step 6: Convert back to x

From $x = 3\sin\theta$ : $\sin\theta = \frac{x}{3}$ , $\theta = \arcsin\frac{x}{3}$

$\cos\theta = \frac{\sqrt{9-x^2}}{3}$

$= \frac{9}{2}\arcsin\frac{x}{3} - \frac{9}{2} \cdot \frac{x}{3} \cdot \frac{\sqrt{9-x^2}}{3} + C$

$= \frac{9}{2}\arcsin\frac{x}{3} - \frac{x\sqrt{9-x^2}}{2} + C$

Answer: $\frac{9}{2}\arcsin\frac{x}{3} - \frac{x\sqrt{9-x^2}}{2} + C$

2Problem 2expert

❓ Question:

Evaluate $\int \frac{dx}{(x^2+1)^2}$ using trigonometric substitution.

💡 Show Solution

Step 1: Identify the form

$x^2 + 1 = x^2 + 1^2$ → Case 2, use $x = \tan\theta$

Step 2: Make substitution

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

$x^2 + 1 = \tan^2\theta + 1 = \sec^2\theta$

$(x^2+1)^2 = \sec^4\theta$

Step 3: Rewrite integral

$\int \frac{dx}{(x^2+1)^2} = \int \frac{\sec^2\theta\,d\theta}{\sec^4\theta}$

$= \int \frac{1}{\sec^2\theta}\,d\theta = \int \cos^2\theta\,d\theta$

Step 4: Use identity $\cos^2\theta = \frac{1+\cos 2\theta}{2}$

$= \int \frac{1+\cos 2\theta}{2}\,d\theta$

$= \frac{1}{2}\left(\theta + \frac{\sin 2\theta}{2}\right) + C$

$= \frac{\theta}{2} + \frac{\sin 2\theta}{4} + C$

Step 5: Use $\sin 2\theta = 2\sin\theta\cos\theta$

$= \frac{\theta}{2} + \frac{2\sin\theta\cos\theta}{4} + C$

$= \frac{\theta}{2} + \frac{\sin\theta\cos\theta}{2} + C$

Step 6: Convert back to x

From $x = \tan\theta$ : $\theta = \arctan x$

$\sin\theta = \frac{x}{\sqrt{1+x^2}}$ , $\cos\theta = \frac{1}{\sqrt{1+x^2}}$

$= \frac{\arctan x}{2} + \frac{1}{2} \cdot \frac{x}{\sqrt{1+x^2}} \cdot \frac{1}{\sqrt{1+x^2}} + C$

$= \frac{\arctan x}{2} + \frac{x}{2(1+x^2)} + C$

Answer: $\frac{\arctan x}{2} + \frac{x}{2(1+x^2)} + C$

3Problem 3hard

❓ Question:

Evaluate $\int \frac{dx}{x\sqrt{x^2-16}}$ using trigonometric substitution.

💡 Show Solution

Step 1: Identify the form

$\sqrt{x^2-16} = \sqrt{x^2 - 4^2}$ → Case 3, use $x = 4\sec\theta$

Step 2: Make substitution

$x = 4\sec\theta$ , $dx = 4\sec\theta\tan\theta\,d\theta$

$\sqrt{x^2-16} = \sqrt{16\sec^2\theta - 16} = 4\tan\theta$

Step 3: Rewrite integral

$\int \frac{dx}{x\sqrt{x^2-16}} = \int \frac{4\sec\theta\tan\theta\,d\theta}{4\sec\theta \cdot 4\tan\theta}$

$= \int \frac{4\sec\theta\tan\theta}{16\sec\theta\tan\theta}\,d\theta$

$= \int \frac{1}{4}\,d\theta = \frac{\theta}{4} + C$

Step 4: Convert back to x

From $x = 4\sec\theta$ : $\sec\theta = \frac{x}{4}$

$\theta = \text{arcsec}\frac{x}{4}$

Or equivalently: $\theta = \arccos\frac{4}{x}$

Answer: $\frac{1}{4}\text{arcsec}\frac{x}{4} + C$ or $\frac{1}{4}\arccos\frac{4}{x} + C$

4Problem 4medium

❓ Question:

Evaluate ∫√(9 - x²) dx using trigonometric substitution.

💡 Show Solution

Step 1: Identify the form: √(9 - x²) = √(a² - x²) where a = 3 Use substitution x = a·sin(θ) = 3sin(θ)

Step 2: Find dx: dx = 3cos(θ) dθ

Step 3: Substitute into √(9 - x²): √(9 - x²) = √(9 - 9sin²(θ)) = √(9(1 - sin²(θ))) = √(9cos²(θ)) = 3|cos(θ)| = 3cos(θ) (assuming -π/2 ≤ θ ≤ π/2)

Step 4: Substitute into integral: ∫√(9 - x²) dx = ∫3cos(θ)·3cos(θ) dθ = ∫9cos²(θ) dθ

Step 5: Use identity cos²(θ) = (1 + cos(2θ))/2: = ∫9·(1 + cos(2θ))/2 dθ = (9/2)∫(1 + cos(2θ)) dθ = (9/2)[θ + sin(2θ)/2] + C

Step 6: Convert back to x: sin(θ) = x/3, so θ = arcsin(x/3) sin(2θ) = 2sin(θ)cos(θ) = 2(x/3)·√(9-x²)/3 = 2x√(9-x²)/9

= (9/2)[arcsin(x/3) + x√(9-x²)/9] + C = (9/2)arcsin(x/3) + (x√(9-x²))/2 + C

Answer: (9/2)arcsin(x/3) + (x√(9-x²))/2 + C

5Problem 5hard

❓ Question:

Evaluate ∫dx/(x²√(x² + 4)) using trig substitution.

💡 Show Solution

Step 1: Identify the form: √(x² + 4) = √(x² + a²) where a = 2 Use substitution x = a·tan(θ) = 2tan(θ)

Step 2: Find dx: dx = 2sec²(θ) dθ

Step 3: Substitute √(x² + 4): √(x² + 4) = √(4tan²(θ) + 4) = √(4(tan²(θ) + 1)) = √(4sec²(θ)) = 2sec(θ)

Step 4: Substitute into integral: ∫dx/(x²√(x² + 4)) = ∫(2sec²(θ) dθ)/((4tan²(θ))·(2sec(θ))) = ∫(2sec²(θ))/(8tan²(θ)sec(θ)) dθ = ∫sec(θ)/(4tan²(θ)) dθ = (1/4)∫(1/cos(θ))·(cos²(θ)/sin²(θ)) dθ = (1/4)∫cos(θ)/sin²(θ) dθ

Step 5: Let u = sin(θ), du = cos(θ) dθ: = (1/4)∫du/u² = (1/4)·(-1/u) + C = -1/(4sin(θ)) + C

Step 6: Convert back to x: tan(θ) = x/2, so sin(θ) = x/√(x² + 4)

= -1/(4·x/√(x² + 4)) + C = -√(x² + 4)/(4x) + C

Answer: -√(x² + 4)/(4x) + C

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Trigonometric Substitution

🔺 Trigonometric Substitution

The Problem

The Three Cases

Case 1: a2−x2\sqrt{a^2 - x^2}a2−x2​

Case 2: a2+x2\sqrt{a^2 + x^2}a2+x2​

Case 3: x2−a2\sqrt{x^2 - a^2}x2−a2​

Quick Reference Table

Example 1: Case 1 - a2−x2\sqrt{a^2 - x^2}a2−x2​

Example 2: Case 2 - a2+x2\sqrt{a^2 + x^2}a2+x2​

Example 3: Case 3 - x2−a2\sqrt{x^2 - a^2}x2−a2​

Drawing the Reference Triangle

For x=asin⁡θx = a\sin\thetax=asinθ:

For x=atan⁡θx = a\tan\thetax=atanθ:

For x=asec⁡θx = a\sec\thetax=asecθ:

When to Use Trig Substitution

Use when you see:

Don't use when:

⚠️ Common Mistakes

Mistake 1: Wrong Substitution

Mistake 2: Forgetting to Convert Back

Mistake 3: Sign Errors

Mistake 4: Forgetting dx

Completing the Square First

Standard Trig Integrals to Know

📝 Practice Strategy

📚 Practice Problems

1Problem 1hard

❓ Question:

2Problem 2expert

❓ Question:

3Problem 3hard

❓ Question:

4Problem 4medium

❓ Question:

5Problem 5hard

❓ Question:

Practice with Flashcards

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Case 1: $\sqrt{a^2 - x^2}$

Case 2: $\sqrt{a^2 + x^2}$

Case 3: $\sqrt{x^2 - a^2}$

Example 1: Case 1 - $\sqrt{a^2 - x^2}$

Example 2: Case 2 - $\sqrt{a^2 + x^2}$

Example 3: Case 3 - $\sqrt{x^2 - a^2}$

For $x = a\sin\theta$ :

For $x = a\tan\theta$ :

For $x = a\sec\theta$ :