Review: Integration Using Trigonometric Substitution

Trigonometric substitution (“trig sub”) is an effective integration technique designed to eliminate radicals of the form \(\sqrt{\pm x^2 \pm a^2}\) by leveraging fundamental Pythagorean trigonometric identities.

1. Summary of the Three Standard Cases

Expression Form Substitution Differential \(dx\) Identity Used Simplified Radical Domain Restriction
Case 1: \(\sqrt{a^2 - x^2}\) \(x = a\sin\theta\) \(dx = a\cos\theta\,d\theta\) \(1 - \sin^2\theta = \cos^2\theta\) \(a\cos\theta\) \(-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}\)
Case 2: \(\sqrt{a^2 + x^2}\) \(x = a\tan\theta\) \(dx = a\sec^2\theta\,d\theta\) \(1 + \tan^2\theta = \sec^2\theta\) \(a\sec\theta\) \(-\frac{\pi}{2} < \theta < \frac{\pi}{2}\)
Case 3: \(\sqrt{x^2 - a^2}\) \(x = a\sec\theta\) \(dx = a\sec\theta\tan\theta\,d\theta\) \(\sec^2\theta - 1 = \tan^2\theta\) \(a\tan\theta\) \(0 \le \theta < \frac{\pi}{2}\) or \(\pi \le \theta < \frac{3\pi}{2}\)

(Assuming \(a > 0\).)

2. Standard Workflow

  1. Identify the radical / quadratic pattern: Determine which form matches: \(a^2 - x^2\), \(a^2 + x^2\), or \(x^2 - a^2\). Complete the square if a linear \(x\)-term is present.
  2. Substitute \(x\) and \(dx\): Replace every instance of \(x\) and \(dx\) with their respective trigonometric expressions.
  3. Simplify the radical: Apply the corresponding Pythagorean identity so the square root simplifies to a single trigonometric term.
  4. Evaluate the resulting trig integral: Use reduction formulas, half-angle identities, or standard trig integral strategies.
  5. Back-substitute to \(x\): Draw a reference right triangle using \(\theta\) and the original substitution to express all trigonometric functions in terms of the original variable \(x\).

3. Detailed Examples

Example 1: Form \(\sqrt{a^2 - x^2}\) (Sine Substitution)

Evaluate: \[\int \frac{\sqrt{9 - x^2}}{x^2}\,dx\]

  • Step 1: Set up substitution
    Here \(a = 3\), so let: \[x = 3\sin\theta \implies dx = 3\cos\theta\,d\theta\] \[\sqrt{9 - x^2} = \sqrt{9(1 - \sin^2\theta)} = 3\cos\theta\]

  • Step 2: Substitute into the integral \[\int \frac{3\cos\theta}{(3\sin\theta)^2} \cdot (3\cos\theta\,d\theta) = \int \frac{9\cos^2\theta}{9\sin^2\theta}\,d\theta = \int \cot^2\theta\,d\theta\]

  • Step 3: Integrate
    Use the identity \(\cot^2\theta = \csc^2\theta - 1\): \[\int (\csc^2\theta - 1)\,d\theta = -\cot\theta - \theta + C\]

  • Step 4: Back-substitute via reference triangle
    Since \(\sin\theta = \frac{x}{3}\):

    • \(\text{Opposite} = x\)
    • \(\text{Hypotenuse} = 3\)
    • \(\text{Adjacent} = \sqrt{9 - x^2}\)
    • \(\cot\theta = \frac{\text{Adjacent}}{\text{Opposite}} = \frac{\sqrt{9 - x^2}}{x}\)
    • \(\theta = \arcsin\left(\frac{x}{3}\right)\)

  • Final Result: \[\int \frac{\sqrt{9 - x^2}}{x^2}\,dx = -\frac{\sqrt{9 - x^2}}{x} - \arcsin\left(\frac{x}{3}\right) + C\]

Example 2: Form \(\sqrt{a^2 + x^2}\) (Tangent Substitution)

Evaluate: \[\int \frac{1}{(x^2 + 4)^{3/2}}\,dx\]

  • Step 1: Set up substitution
    Here \(a = 2\), so let: \[x = 2\tan\theta \implies dx = 2\sec^2\theta\,d\theta\] \[(x^2 + 4)^{3/2} = (4\tan^2\theta + 4)^{3/2} = (4\sec^2\theta)^{3/2} = 8\sec^3\theta\]

  • Step 2: Substitute and simplify \[\int \frac{2\sec^2\theta\,d\theta}{8\sec^3\theta} = \frac{1}{4}\int \frac{1}{\sec\theta}\,d\theta = \frac{1}{4}\int \cos\theta\,d\theta\]

  • Step 3: Integrate \[\frac{1}{4}\int \cos\theta\,d\theta = \frac{1}{4}\sin\theta + C\]

  • Step 4: Back-substitute via reference triangle
    Since \(\tan\theta = \frac{x}{2}\):

    • \(\text{Opposite} = x\)
    • \(\text{Adjacent} = 2\)
    • \(\text{Hypotenuse} = \sqrt{x^2 + 4}\)
    • \(\sin\theta = \frac{x}{\sqrt{x^2 + 4}}\)

  • Final Result: \[\int \frac{1}{(x^2 + 4)^{3/2}}\,dx = \frac{x}{4\sqrt{x^2 + 4}} + C\]

Example 3: Form \(\sqrt{x^2 - a^2}\) (Secant Substitution)

Evaluate: \[\int \frac{1}{x\sqrt{x^2 - 16}}\,dx \quad (\text{for } x > 4)\]

  • Step 1: Set up substitution
    Here \(a = 4\), so let: \[x = 4\sec\theta \implies dx = 4\sec\theta\tan\theta\,d\theta\] \[\sqrt{x^2 - 16} = \sqrt{16(\sec^2\theta - 1)} = 4\tan\theta\]

  • Step 2: Substitute and simplify \[\int \frac{4\sec\theta\tan\theta\,d\theta}{(4\sec\theta)(4\tan\theta)} = \frac{1}{4}\int 1\,d\theta = \frac{1}{4}\theta + C\]

  • Step 3: Back-substitute
    Since \(\sec\theta = \frac{x}{4} \implies \theta = \operatorname{arcsec}\left(\frac{x}{4}\right)\) (or \(\arccos\left(\frac{4}{x}\right)\)):

  • Final Result: \[\int \frac{1}{x\sqrt{x^2 - 16}}\,dx = \frac{1}{4}\operatorname{arcsec}\left(\frac{x}{4}\right) + C\]

4. Key Pitfalls & Quick Tips

  1. Always substitute the differential \(dx\): A common mistake is leaving \(dx\) as \(d\theta\) without the derivative coefficient (e.g., forgetting the factor \(a\cos\theta\) or \(a\sec^2\theta\)).
  2. Do not force trig sub: Check for standard \(u\)-substitution first. For instance, \(\int x\sqrt{a^2 - x^2}\,dx\) can be solved directly via \(u = a^2 - x^2\) without trigonometry.
  3. Definite integrals: If boundaries are provided, converting limits directly to \(\theta\) avoids constructing reference triangles for back-substitution.