🔺 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!

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

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

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

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Quick Reference Table

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

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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!

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Step 2: Make substitution

Let $x = 2\sin\theta$

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

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

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Step 3: Rewrite integral

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

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

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

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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!

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Step 2: Make substitution

Let $x = \tan\theta$

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

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

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Step 3: Rewrite integral

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

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

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

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

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

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Step 5: Integrate

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

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

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

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

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

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

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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!

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Step 2: Make substitution

Let $x = 3\sec\theta$

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

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

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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!

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

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

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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)

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

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For $x = a\tan\theta$:

Triangle with:

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

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

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For $x = a\sec\theta$:

Triangle with:

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

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

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When to Use Trig Substitution

Use when you see:

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

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⚠️ Common Mistakes

Mistake 1: Wrong Substitution

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

Match the pattern carefully!

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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.

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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.

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Mistake 4: Forgetting dx

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

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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 $a = 1$.

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Standard Trig Integrals to Know

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

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

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

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

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📝 Practice Strategy

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