3.3 Trigonometric Substitution

Trigonometric substitution transforms integrals containing square roots of quadratic expressions into trigonometric integrals that are easier to evaluate. It works by exploiting the Pythagorean identities to eliminate square roots entirely.

Trigonometric Substitution

Trigonometric substitution for square roots

This technique applies to integrals containing expressions of the form $\sqrt{ax^2 + bx + c}$. The core idea: if you can match the expression under the square root to one of three standard forms, you can substitute a trig function for $x$ that causes the square root to collapse via a Pythagorean identity.

The three standard forms are:

• $\sqrt{a^2 - x^2}$ (difference form)
• $\sqrt{a^2 + x^2}$ (sum form)
• $\sqrt{x^2 - a^2}$ (reverse difference form)

If your integrand doesn't immediately match one of these, you may need to complete the square first to get it into the right shape.

Choosing appropriate substitutions

Each form has a specific substitution paired with the Pythagorean identity that eliminates the square root.

When you see $\sqrt{a^2 - x^2}$, substitute $x = a\sin\theta$:

• Then $dx = a\cos\theta \, d\theta$
• The square root simplifies: $\sqrt{a^2 - a^2\sin^2\theta} = a\sqrt{1 - \sin^2\theta} = a\cos\theta$
• Restrict $\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ so that $\cos\theta \geq 0$
• Example: For $\int \sqrt{9 - x^2} \, dx$, identify $a = 3$ and set $x = 3\sin\theta$

When you see $\sqrt{a^2 + x^2}$, substitute $x = a\tan\theta$:

• Then $dx = a\sec^2\theta \, d\theta$
• The square root simplifies: $\sqrt{a^2 + a^2\tan^2\theta} = a\sqrt{1 + \tan^2\theta} = a\sec\theta$
• Restrict $\theta \in (-\frac{\pi}{2}, \frac{\pi}{2})$ so that $\sec\theta > 0$
• Example: For $\int \frac{x}{\sqrt{4 + x^2}} \, dx$, identify $a = 2$ and set $x = 2\tan\theta$

When you see $\sqrt{x^2 - a^2}$, substitute $x = a\sec\theta$:

• Then $dx = a\sec\theta\tan\theta \, d\theta$
• The square root simplifies: $\sqrt{a^2\sec^2\theta - a^2} = a\sqrt{\sec^2\theta - 1} = a\tan\theta$
• Restrict $\theta \in [0, \frac{\pi}{2})$ for $x > a$ or $\theta \in (\frac{\pi}{2}, \pi]$ for $x < -a$, so that $\tan\theta$ has the correct sign
• Example: For $\int \sqrt{x^2 - 1} \, dx$, identify $a = 1$ and set $x = \sec\theta$

Converting solutions back to $x$

After integrating in $\theta$, you need to rewrite everything in terms of $x$. Here's a reliable process:

1. Solve for $\theta$ using the inverse of your substitution:

   • $x = a\sin\theta \implies \theta = \arcsin\!\left(\frac{x}{a}\right)$
   • $x = a\tan\theta \implies \theta = \arctan\!\left(\frac{x}{a}\right)$
   • $x = a\sec\theta \implies \theta = \text{arcsec}\!\left(\frac{x}{a}\right)$

2. Draw a reference right triangle labeled with the substitution. For instance, if $x = a\tan\theta$, the triangle has opposite side $x$, adjacent side $a$, and hypotenuse $\sqrt{a^2 + x^2}$. This lets you read off any trig function of $\theta$ directly in terms of $x$.

3. Replace all trig functions of $\theta$ in your answer using the triangle, and simplify.

Example: Suppose after substituting $x = 3\tan\theta$ you get $2\theta + \sin\theta\cos\theta + C$. From the triangle: $\sin\theta = \frac{x}{\sqrt{9+x^2}}$ and $\cos\theta = \frac{3}{\sqrt{9+x^2}}$. So the final answer is:

$2\arctan\!\left(\frac{x}{3}\right) + \frac{3x}{9+x^2} + C$

The reference triangle step is the most reliable way to handle the back-substitution. Trying to convert without it often leads to sign errors or missed simplifications.

Related techniques and applications

• Completing the square is often a necessary first step. An expression like $\sqrt{-x^2 + 6x - 5}$ doesn't match any standard form until you rewrite it as $\sqrt{4 - (x-3)^2}$, then substitute $x - 3 = 2\sin\theta$.
• Integration by parts sometimes appears after the trig substitution, particularly with the $\sec\theta$ substitution (e.g., $\int \sec^3\theta \, d\theta$ requires integration by parts).
• Hyperbolic substitutions (e.g., $x = a\sinh t$) can replace trig substitutions in some cases and avoid dealing with secant integrals, though they're less commonly tested.
• Partial fractions may be needed if, after substitution, the integrand becomes a rational function of trig functions.