Integrals Yielding Inverse Trigonometric Functions

4 min read

Reference: Stewart §6.6

When Integrals Produce Angles

You’ve learned that $\frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1-x^2}}$. Reading this backwards: $$\int \frac{1}{\sqrt{1-x^2}} \, dx = \arcsin x + C$$

This is remarkable: an algebraic integrand produces a transcendental antiderivative!

The key to this skill is recognizing when an integral has the right form, then using substitution to match it exactly.

📋 Before You Start: Check Your Prerequisites

Inverse trig derivatives: What is $\frac{d}{dx}[\arcsin(x)]$? What is $\frac{d}{dx}[\arctan(x)]$?
- If unsure: Review Derivatives of Inverse Trig
u-substitution: Evaluate $\int 2x \cdot e^{x^2} \, dx$.
- If unsure: Review u-Substitution
Pattern recognition: Does $\int \frac{x}{1+x^2} \, dx$ yield $\arctan$ or $\ln$?
- Answer: $\frac{1}{2}\ln(1+x^2) + C$ (the $x$ in the numerator means we use $u = 1+x^2$)

Prerequisite Map

Quick Reference

Property	Value
Concept	Inverse Functions
Difficulty	Intermediate
Time	~25 minutes

Key Concepts

The Two Essential Formulas

Basic Integration Formulas

$$\int \frac{1}{\sqrt{1-x^2}} \, dx = \arcsin(x) + C$$

$$\int \frac{1}{1+x^2} \, dx = \arctan(x) + C$$

The Generalized Formulas

When the integrand has the form $\frac{1}{\sqrt{a^2-x^2}}$ or $\frac{1}{a^2+x^2}$, use these:

Generalized Formulas (with parameter $a > 0$)

$$\int \frac{1}{\sqrt{a^2-x^2}} \, dx = \arcsin\left(\frac{x}{a}\right) + C$$

$$\int \frac{1}{a^2+x^2} \, dx = \frac{1}{a}\arctan\left(\frac{x}{a}\right) + C$$

Deriving the Generalized Arctangent Formula

Derive $\int \frac{1}{x^2+a^2} \, dx$ using substitution.

Step 1: Factor out $a^2$ to create the standard form: $$\int \frac{dx}{x^2+a^2} = \int \frac{dx}{a^2\left(\frac{x^2}{a^2}+1\right)} = \frac{1}{a^2} \int \frac{dx}{\left(\frac{x}{a}\right)^2+1}$$

Step 2: Substitute $u = x/a$, so $du = dx/a$ and $dx = a \, du$: $$\frac{1}{a^2} \int \frac{a \, du}{u^2+1} = \frac{1}{a} \int \frac{du}{u^2+1} = \frac{1}{a} \arctan(u) + C$$

Step 3: Substitute back: $$\int \frac{1}{x^2+a^2} \, dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C$$

Recognizing the Patterns

Integrand Form	Antiderivative	Key Recognition
$\frac{1}{\sqrt{1-u^2}}$	$\arcsin(u) + C$	Square root of $(1 - \text{something}^2)$
$\frac{1}{\sqrt{a^2-u^2}}$	$\arcsin(u/a) + C$	Square root of $(a^2 - \text{something}^2)$
$\frac{1}{1+u^2}$	$\arctan(u) + C$	$(1 + \text{something}^2)$ in denominator
$\frac{1}{a^2+u^2}$	$\frac{1}{a}\arctan(u/a) + C$	$(a^2 + \text{something}^2)$ in denominator

Warning: Similar-Looking Integrals That DON’T Yield Inverse Trig

Not every integral with $\sqrt{1-x^2}$ or $1+x^2$ produces inverse trig!

Integral	Method	Result
$\int \frac{x}{\sqrt{1-x^2}} \, dx$	Sub $u=1-x^2$	$-\sqrt{1-x^2}+C$
$\int \frac{x}{1+x^2} \, dx$	Sub $u=1+x^2$	$\frac{1}{2}\ln(1+x^2)+C$
$\int \frac{1}{\sqrt{1-x^2}} \, dx$	Inverse trig	$\arcsin(x)+C$
$\int \frac{1}{1+x^2} \, dx$	Inverse trig	$\arctan(x)+C$

The difference: If there’s an $x$ in the numerator, try u-substitution first!

Strategy for These Integrals

Look for $\sqrt{a^2-(\text{expr})^2}$ or $a^2+(\text{expr})^2$ in the denominator
Check the numerator: If it contains the derivative of the expression, use regular u-substitution
If numerator is constant: Use the inverse trig formulas
Manipulate to match the form: Complete the square if needed, factor out constants

Practice Problems

Level 1 Direct Application of Basic Formula

Evaluate $\displaystyle\int \frac{1}{1+x^2} \, dx$.

Thought Process

This matches the standard form $\int \frac{1}{1+x^2} \, dx$ exactly.

We recognize this as the derivative of $\arctan(x)$.

Show Answer

$$\int \frac{1}{1+x^2} \, dx = \arctan(x) + C$$

Level 2 Simple Chain Rule Reversal

Evaluate $\displaystyle\int \frac{3}{\sqrt{1-9x^2}} \, dx$.

Thought Process

Rewrite: $\sqrt{1-9x^2} = \sqrt{1-(3x)^2}$.

This looks like the arcsine pattern with $u = 3x$.

Let $u = 3x$, so $du = 3 \, dx$. The “3” in the numerator is exactly what we need!

Show Answer

Let $u = 3x$, so $du = 3 \, dx$.

$$\int \frac{3}{\sqrt{1-(3x)^2}} \, dx = \int \frac{du}{\sqrt{1-u^2}} = \arcsin(u) + C = \arcsin(3x) + C$$

Level 3 Using the Generalized Arctangent Formula

Evaluate $\displaystyle\int \frac{1}{x^2+9} \, dx$.

Thought Process

This has the form $\frac{1}{x^2+a^2}$ with $a = 3$ (since $9 = 3^2$).

Use the formula: $\int \frac{1}{x^2+a^2} \, dx = \frac{1}{a}\arctan\left(\frac{x}{a}\right) + C$

Show Answer

Using $\int \frac{1}{x^2+a^2} \, dx = \frac{1}{a}\arctan\left(\frac{x}{a}\right) + C$ with $a = 3$:

$$\int \frac{1}{x^2+9} \, dx = \frac{1}{3}\arctan\left(\frac{x}{3}\right) + C$$

Level 4 Definite Integral with Substitution

Evaluate $\displaystyle\int_0^{1/3} \frac{1}{\sqrt{1-9x^2}} \, dx$.

Thought Process

Rewrite the square root to see the arcsine pattern: $\sqrt{1-9x^2} = \sqrt{1-(3x)^2}$.

This suggests the substitution $u = 3x$ because then the denominator becomes $\sqrt{1-u^2}$.

With $u = 3x$, we have $du = 3 \, dx$, so $dx = \frac{1}{3} du$.

Change the limits: when $x = 0$, $u = 0$; when $x = 1/3$, $u = 1$.

The upper limit $u = 1$ is at the edge of the domain of arcsine, which means we are integrating “all the way” from the center to the boundary.

Show Answer

Let $u = 3x$, so $du = 3 \, dx$ and $dx = \frac{1}{3} du$.

Change limits: $x = 0 \Rightarrow u = 0$; $x = 1/3 \Rightarrow u = 1$.

$$\int_0^{1/3} \frac{1}{\sqrt{1-(3x)^2}} \, dx = \frac{1}{3}\int_0^{1} \frac{du}{\sqrt{1-u^2}}$$

$$= \frac{1}{3} \left[\arcsin(u)\right]_0^{1} = \frac{1}{3}\left[\arcsin(1) - \arcsin(0)\right]$$

$$= \frac{1}{3}\left[\frac{\pi}{2} - 0\right] = \frac{\pi}{6}$$

Physical interpretation: We integrated from $x=0$ to $x=1/3$, which corresponds to $u=0$ to $u=1$. Since $u=1$ is the maximum of the arcsine domain, we covered the full “half-arc” from $0$ to $\pi/2$, then divided by 3 for the chain rule factor.

Level 5 Deriving the Generalized Arcsine Formula

Prove that for $a > 0$: $$\int \frac{1}{\sqrt{a^2-x^2}} \, dx = \arcsin\left(\frac{x}{a}\right) + C$$

Thought Process

Strategy: Use substitution $u = x/a$ to reduce this to the basic formula $\int \frac{1}{\sqrt{1-u^2}} \, du$.

Key step: Factor $a^2$ out of the square root to get $\sqrt{a^2(1 - x^2/a^2)} = a\sqrt{1 - (x/a)^2}$.

Show Answer

Proof:

Step 1: Rewrite the integrand by factoring $a^2$ from under the square root: $$\sqrt{a^2 - x^2} = \sqrt{a^2\left(1 - \frac{x^2}{a^2}\right)} = a\sqrt{1 - \left(\frac{x}{a}\right)^2}$$

So the integral becomes: $$\int \frac{1}{a\sqrt{1 - (x/a)^2}} \, dx = \frac{1}{a}\int \frac{dx}{\sqrt{1 - (x/a)^2}}$$

Step 2: Substitute $u = x/a$, so $du = \frac{1}{a}dx$ and $dx = a \, du$: $$\frac{1}{a}\int \frac{a \, du}{\sqrt{1 - u^2}} = \int \frac{du}{\sqrt{1-u^2}} = \arcsin(u) + C$$

Step 3: Substitute back: $$\int \frac{1}{\sqrt{a^2-x^2}} \, dx = \arcsin\left(\frac{x}{a}\right) + C \quad \square$$

Conceptual Check (CCI-Style)

Question: A student evaluates $\int \frac{x}{1+x^2} \, dx$ and gets $\arctan(x) + C$. Is this correct?

(A) Yes, this is correct (B) No; the answer should be $\frac{1}{2}\ln(1+x^2) + C$ (C) No; the answer should be $\frac{1}{2}\arctan(x^2) + C$ (D) No; this integral cannot be evaluated with elementary functions

Answer

(B) When there’s an $x$ in the numerator, use u-substitution with $u = 1+x^2$: $$\int \frac{x}{1+x^2} \, dx = \frac{1}{2}\int \frac{du}{u} = \frac{1}{2}\ln\vert u\vert + C = \frac{1}{2}\ln(1+x^2) + C$$

The arctangent formula only applies when the numerator is a constant (or can be made into one after substitution).

Common Mistakes to Avoid

Mistake	Correction
Using $\arctan$ for $\int \frac{x}{1+x^2} dx$	Check for $x$ in numerator → use ln
Forgetting the $\frac{1}{a}$ factor in $\int \frac{1}{x^2+a^2} dx$	Formula gives $\frac{1}{a}\arctan(x/a)$, not $\arctan(x/a)$
Confusing $\int \frac{1}{\sqrt{1-x^2}} dx$ with $\int \frac{1}{\sqrt{x^2-1}} dx$	First is $\arcsin$; second requires $\text{arcsec}$ or hyperbolic
Writing $\arcsin(x^2) + C$ for $\int \frac{2x}{\sqrt{1-x^4}} dx$	This IS correct! (Let $u = x^2$)

Common Misconceptions

Common misconception

$\int \frac{1}{a^2 + x^2}\,dx = \arctan(x) + C$, omitting the factor $\frac{1}{a}$.

This is the composition-is-not-chaining error. The general formula is $\int \frac{1}{a^2 + x^2}\,dx = \frac{1}{a}\arctan\!\left(\frac{x}{a}\right) + C$. The factor $\frac{1}{a}$ appears because the correct substitution $u = \frac{x}{a}$ introduces $du = \frac{dx}{a}$, contributing that constant. Verification by differentiation confirms the need: $\frac{d}{dx}\!\left[\frac{1}{a}\arctan\!\left(\frac{x}{a}\right)\right] = \frac{1}{a} \cdot \frac{1/a}{1 + x^2/a^2} = \frac{1}{a^2 + x^2}$.

Mastery Checklist

I can state the basic formulas for $\int \frac{1}{\sqrt{1-x^2}} dx$ and $\int \frac{1}{1+x^2} dx$
I can apply the generalized formulas with parameter $a$
I can distinguish between integrals that yield inverse trig vs. logarithms
I can use substitution to convert integrals to standard inverse trig forms
I can evaluate definite integrals involving inverse trig functions

Mental Model

Pattern Matching: Think of these formulas as “templates” to match:

Template 1: $\frac{\text{constant}}{\sqrt{a^2 - (\text{stuff})^2}}$ → arcsine
Template 2: $\frac{\text{constant}}{a^2 + (\text{stuff})^2}$ → arctangent

The “stuff” becomes the argument of the inverse trig function, and substitution handles the chain rule.

Connections

Looking back:

Derivative formulas read backwards give these integration formulas
u-substitution is essential for matching the standard forms

Looking ahead:

Trig substitution handles integrals with $\sqrt{x^2 + a^2}$ and $\sqrt{x^2 - a^2}$
These patterns appear in many applications including arc length calculations

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Derivatives of Inverse Trig	Skills Index	Hyperbolic Functions

Last updated: 2026-01-22