Trig Substitutions for MATH 141
Exam Relevance for MATH 141
Trig sub is heavily tested in MATH 141. Know which substitution for each radical form—this appears on every exam.
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.