Trig Substitutions for MATH 101 A
This skill appears on:
Understanding Trigonometric Substitution
Some integrals contain expressions like $\sqrt{a^2 - x^2}$, $\sqrt{a^2 + x^2}$, or $\sqrt{x^2 - a^2}$ that can't be handled with u-substitution or basic techniques. These square roots are stubborn—they don't simplify easily.
Trigonometric substitution is the solution. By replacing $x$ with a trig expression, we can use Pythagorean identities to eliminate the square root entirely!
The Three Cases
Each type of square root has a specific substitution:
| Expression | Substitution | Why It Works |
|---|---|---|
| $\sqrt{a^2 - x^2}$ | $x = a\sin\theta$ | $a^2 - a^2\sin^2\theta = a^2\cos^2\theta$ |
| $\sqrt{a^2 + x^2}$ | $x = a\tan\theta$ | $a^2 + a^2\tan^2\theta = a^2\sec^2\theta$ |
| $\sqrt{x^2 - a^2}$ | $x = a\sec\theta$ | $a^2\sec^2\theta - a^2 = a^2\tan^2\theta$ |
Memory trick: Look at what's under the square root:
- Minus $x^2$ → use sine (both have "minus" energy—sine is bounded)
- Plus $x^2$ → use tangent
- $x^2$ minus → use secant
Key Steps for Every Problem
- Identify which case you have ($a^2 - x^2$, $a^2 + x^2$, or $x^2 - a^2$)
- Substitute $x$ with the appropriate trig expression
- Find $dx$ by differentiating your substitution
- Simplify the square root using trig identities
- Integrate the resulting trig integral
- Convert back to $x$ using a reference triangle
Problem: Find $\int \sqrt{16 - x^2} \, dx$
Step 1: Identify the case
We have $\sqrt{16 - x^2} = \sqrt{4^2 - x^2}$
This is the $\sqrt{a^2 - x^2}$ case with $a = 4$.
Step 2: Substitute
Let $x = 4\sin\theta$
Then $dx = 4\cos\theta \, d\theta$
Step 3: Simplify the square root
$$\sqrt{16 - x^2} = \sqrt{16 - 16\sin^2\theta}$$
$$= \sqrt{16(1 - \sin^2\theta)}$$
$$= \sqrt{16\cos^2\theta} = 4\cos\theta$$
Step 4: Rewrite the integral
$$\int \sqrt{16 - x^2} \, dx$$
$$= \int 4\cos\theta \cdot 4\cos\theta \, d\theta$$
$$= 16\int \cos^2\theta \, d\theta$$
Step 5: Integrate using power-reducing
$$\cos^2\theta = \frac{1 + \cos 2\theta}{2}$$
$$= 16 \cdot \frac{1}{2}\int (1 + \cos 2\theta) \, d\theta$$
$$= 8\left(\theta + \frac{\sin 2\theta}{2}\right) + C$$
$$= 8\theta + 4\sin 2\theta + C$$
Step 6: Use double angle identity
$\sin 2\theta = 2\sin\theta\cos\theta$
$$= 8\theta + 8\sin\theta\cos\theta + C$$
Step 7: Convert back to $x$
From $x = 4\sin\theta$: $\sin\theta = \frac{x}{4}$, so $\theta = \arcsin\frac{x}{4}$
For $\cos\theta$: draw a triangle with $\sin\theta = \frac{x}{4}$
$$\cos\theta = \frac{\sqrt{16-x^2}}{4}$$
Substituting:
$$= 8\arcsin\frac{x}{4} + 8 \cdot \frac{x}{4} \cdot \frac{\sqrt{16-x^2}}{4} + C$$
$$= \boxed{8\arcsin\frac{x}{4} + \frac{x\sqrt{16-x^2}}{2} + C}$$
Problem: Find $\int \frac{1}{\sqrt{9 + x^2}} \, dx$
Step 1: Identify the case
We have $\sqrt{9 + x^2} = \sqrt{3^2 + x^2}$
This is the $\sqrt{a^2 + x^2}$ case with $a = 3$.
Step 2: Substitute
Let $x = 3\tan\theta$
Then $dx = 3\sec^2\theta \, d\theta$
Step 3: Simplify the square root
$$\sqrt{9 + x^2} = \sqrt{9 + 9\tan^2\theta}$$
$$= \sqrt{9(1 + \tan^2\theta)}$$
$$= \sqrt{9\sec^2\theta} = 3\sec\theta$$
Step 4: Rewrite and integrate
$$\int \frac{1}{\sqrt{9 + x^2}} \, dx$$
$$= \int \frac{3\sec^2\theta}{3\sec\theta} \, d\theta$$
$$= \int \sec\theta \, d\theta$$
$$= \ln|\sec\theta + \tan\theta| + C$$
Step 5: Convert back to $x$
From $x = 3\tan\theta$: $\tan\theta = \frac{x}{3}$
Draw triangle: $\sec\theta = \frac{\sqrt{9+x^2}}{3}$
$$= \ln\left|\frac{\sqrt{9+x^2}}{3} + \frac{x}{3}\right| + C$$
$$= \boxed{\ln\left|\sqrt{9+x^2} + x\right| + C_1}$$
(We absorbed $-\ln 3$ into the constant)
Problem: Find $\int \frac{\sqrt{x^2 - 4}}{x} \, dx$
Step 1: Identify the case
We have $\sqrt{x^2 - 4} = \sqrt{x^2 - 2^2}$
This is the $\sqrt{x^2 - a^2}$ case with $a = 2$.
Step 2: Substitute
Let $x = 2\sec\theta$
Then $dx = 2\sec\theta\tan\theta \, d\theta$
Step 3: Simplify the square root
$$\sqrt{x^2 - 4} = \sqrt{4\sec^2\theta - 4}$$
$$= \sqrt{4(\sec^2\theta - 1)}$$
$$= \sqrt{4\tan^2\theta} = 2\tan\theta$$
Step 4: Rewrite the integral
$$\int \frac{\sqrt{x^2-4}}{x} \, dx$$
$$= \int \frac{2\tan\theta}{2\sec\theta} \cdot 2\sec\theta\tan\theta \, d\theta$$
$$= \int 2\tan^2\theta \, d\theta$$
Step 5: Integrate
Use $\tan^2\theta = \sec^2\theta - 1$:
$$= 2\int (\sec^2\theta - 1) \, d\theta$$
$$= 2(\tan\theta - \theta) + C$$
Step 6: Convert back to $x$
From $x = 2\sec\theta$: $\sec\theta = \frac{x}{2}$
So $\theta = \text{arcsec}\frac{x}{2}$
Draw triangle: $\tan\theta = \frac{\sqrt{x^2-4}}{2}$
$$= 2 \cdot \frac{\sqrt{x^2-4}}{2} - 2\text{arcsec}\frac{x}{2} + C$$
$$= \boxed{\sqrt{x^2-4} - 2\text{arcsec}\frac{x}{2} + C}$$
Problem: Find $\int \frac{1}{\sqrt{x^2 + 6x + 13}} \, dx$
Step 1: Complete the square
$$x^2 + 6x + 13 = (x^2 + 6x + 9) + 4$$
$$= (x + 3)^2 + 4$$
Step 2: Substitute to simplify
Let $u = x + 3$, so $du = dx$
$$\int \frac{1}{\sqrt{u^2 + 4}} \, du$$
This is now $\sqrt{u^2 + a^2}$ with $a = 2$.
Step 3: Trig substitution
Let $u = 2\tan\theta$, so $du = 2\sec^2\theta \, d\theta$
$$\sqrt{u^2 + 4} = 2\sec\theta$$
Step 4: Integrate
$$= \int \frac{2\sec^2\theta}{2\sec\theta} \, d\theta = \int \sec\theta \, d\theta$$
$$= \ln|\sec\theta + \tan\theta| + C$$
Step 5: Convert back
$\tan\theta = \frac{u}{2}$, $\sec\theta = \frac{\sqrt{u^2+4}}{2}$
$$= \ln\left|\frac{\sqrt{u^2+4} + u}{2}\right| + C$$
$$= \ln\left|\sqrt{(x+3)^2+4} + (x+3)\right| + C_1$$
$$= \boxed{\ln\left|\sqrt{x^2+6x+13} + x + 3\right| + C}$$
Drawing Reference Triangles
After integrating, you'll have an answer in terms of $\theta$, but you need the final answer in terms of $x$. A reference triangle helps you convert trig functions of $\theta$ back to expressions with $x$.
Case 1: Converting back from $x = a\sin\theta$
From the substitution: $\sin\theta = \frac{x}{a}$
Draw a right triangle where:
- Opposite = $x$ (numerator of sine)
- Hypotenuse = $a$ (denominator of sine)
- Adjacent = $\sqrt{a^2 - x^2}$ (by Pythagorean theorem)
Mini-example: If $x = 4\sin\theta$, find $\cos\theta$ in terms of $x$.
From $\sin\theta = \frac{x}{4}$: opposite = $x$, hypotenuse = $4$
Adjacent $= \sqrt{16 - x^2}$
So $\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\sqrt{16-x^2}}{4}$
Case 2: Converting back from $x = a\tan\theta$
From the substitution: $\tan\theta = \frac{x}{a}$
Draw a right triangle where:
- Opposite = $x$ (numerator of tangent)
- Adjacent = $a$ (denominator of tangent)
- Hypotenuse = $\sqrt{a^2 + x^2}$ (by Pythagorean theorem)
Mini-example: If $x = 3\tan\theta$, find $\sec\theta$ in terms of $x$.
From $\tan\theta = \frac{x}{3}$: opposite = $x$, adjacent = $3$
Hypotenuse $= \sqrt{9 + x^2}$
So $\sec\theta = \frac{\text{hypotenuse}}{\text{adjacent}} = \frac{\sqrt{9+x^2}}{3}$
Case 3: Converting back from $x = a\sec\theta$
From the substitution: $\sec\theta = \frac{x}{a}$
Draw a right triangle where:
- Hypotenuse = $x$ (numerator of secant)
- Adjacent = $a$ (denominator of secant)
- Opposite = $\sqrt{x^2 - a^2}$ (by Pythagorean theorem)
Mini-example: If $x = 2\sec\theta$, find $\tan\theta$ in terms of $x$.
From $\sec\theta = \frac{x}{2}$: hypotenuse = $x$, adjacent = $2$
Opposite $= \sqrt{x^2 - 4}$
So $\tan\theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{\sqrt{x^2-4}}{2}$
Common Mistakes and Misunderstandings
❌ Mistake: Using the wrong substitution
Wrong: For $\sqrt{9 - x^2}$, using $x = 3\tan\theta$
Why it's wrong: This gives $\sqrt{9 - 9\tan^2\theta}$, which doesn't simplify nicely.
Correct: Use $x = 3\sin\theta$ for $\sqrt{a^2 - x^2}$ forms.
❌ Mistake: Forgetting to find $dx$
Wrong: Substituting $x = 2\sin\theta$ but leaving $dx$ unchanged
Why it's wrong: If $x = 2\sin\theta$, then $dx = 2\cos\theta \, d\theta$, not just $d\theta$.
Correct: Always differentiate your substitution to find $dx$.
❌ Mistake: Not completing the square when needed
Wrong: Trying trig sub directly on $\sqrt{x^2 + 4x + 5}$
Why it's wrong: This doesn't match any standard form.
Correct: Complete the square first:
$$x^2 + 4x + 5 = (x+2)^2 + 1$$
Now use $u = x + 2$, then trig sub on $\sqrt{u^2 + 1}$.
Trig Substitution: √(a² - x²)
Use when the integrand contains √(a² - x²). This makes the square root become a·cos(θ).
Variables:
- $x$:
- the variable being replaced
- $a$:
- the constant in √(a² - x²)
- $\theta$:
- the new variable
- $dx$:
- becomes a·cos(θ) dθ
Trig Substitution: √(a² + x²)
Use when the integrand contains √(a² + x²). This makes the square root become a·sec(θ).
Variables:
- $x$:
- the variable being replaced
- $a$:
- the constant in √(a² + x²)
- $\theta$:
- the new variable
- $dx$:
- becomes a·sec²(θ) dθ
Trig Substitution: √(x² - a²)
Use when the integrand contains √(x² - a²). This makes the square root become a·tan(θ).
Variables:
- $x$:
- the variable being replaced
- $a$:
- the constant in √(x² - a²)
- $\theta$:
- the new variable
- $dx$:
- becomes a·sec(θ)·tan(θ) dθ
This skill is taught in the following courses. Create an account to access practice exercises and full course materials.