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Section 6.6: Inverse Trigonometric Functions

What Is This Section About?

You know that $\sin(\pi/6) = 1/2$. But what if someone asks: "What angle has a sine of $1/2$?"

This question has infinitely many answers! To create inverse trigonometric functions, we must restrict the domains of sine, cosine, and tangent so they become one-to-one. This section covers:

Definitions: How we restrict domains to create $\arcsin$, $\arccos$, and $\arctan$
Derivatives: Formulas like $\frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1-x^2}}$
Integrals: Recognizing when $\int \frac{1}{\sqrt{1-x^2}} dx$ produces inverse trig

Why Should I Care?

Inverse trig functions appear everywhere in calculus and beyond:

Application	Where It Shows Up
Integration	$\int \frac{1}{1+x^2} dx = \arctan(x) + C$
Trig Substitution (Ch. 7)	When you see $\sqrt{1-x^2}$, let $x = \sin\theta$
Physics	Calculating angles from coordinate ratios
Engineering	Phase angles in AC circuits

Skill Dependency Map

graph TD
    subgraph Prerequisites
        A["Trig Functions<br/>(standard angles)"]
        B["Inverse Functions<br/>(basics)"]
        C["Implicit<br/>Differentiation"]
        D["u-Substitution"]
    end

    subgraph "Section 6.6 Skills"
        E["Inverse Trig<br/>Definitions"]
        F["Simplifying<br/>Compositions"]
        G["Derivatives of<br/>Inverse Trig"]
        H["Integrals Yielding<br/>Inverse Trig"]
    end

    subgraph "What This Unlocks"
        I["Trig Substitution<br/>(Ch. 7)"]
        J["Integration<br/>Strategy"]
    end

    A --> E
    B --> E
    E --> F
    E --> G
    C --> G
    F --> G
    G --> H
    D --> H
    H --> I
    H --> J

    style E fill:#d1fae5,stroke:#059669,stroke-width:2px
    style F fill:#d1fae5,stroke:#059669,stroke-width:2px
    style G fill:#d1fae5,stroke:#059669,stroke-width:2px
    style H fill:#d1fae5,stroke:#059669,stroke-width:2px

    click E "inverse-trig-definitions.html"
    click F "simplifying-inverse-trig-expressions.html"
    click G "inverse-trig-derivatives.html"
    click H "integrals-yielding-inverse-trig.html"

    click B "../ch6-sec1/inverse-functions-basics.html"
    click C "../ch2-sec6/implicit-differentiation.html"
    click D "../ch4-sec5/u-substitution-definite.html"
    click J "../ch7-sec5/integration-strategy.html"

Skills in This Section

Skill	What You'll Learn	Difficulty
Inverse Trig Definitions	Domains, ranges, and exact values of $\arcsin$, $\arccos$, $\arctan$	⭐ Beginner
Simplifying Inverse Trig Expressions	The triangle method for expressions like $\tan(\arcsin x)$	⭐⭐ Intermediate
Derivatives of Inverse Trig	Formulas and derivations using implicit differentiation	⭐⭐ Intermediate
Integrals Yielding Inverse Trig	Recognizing and evaluating $\int \frac{1}{\sqrt{a^2-x^2}} dx$ and $\int \frac{1}{a^2+x^2} dx$	⭐⭐ Intermediate

Recommended order: Work through the skills in the order listed above; each builds on the previous.

Quick Reference: Key Formulas

Derivatives

Function	Derivative	Domain
$\arcsin x$	$\frac{1}{\sqrt{1-x^2}}$	$(-1, 1)$
$\arccos x$	$-\frac{1}{\sqrt{1-x^2}}$	$(-1, 1)$
$\arctan x$	$\frac{1}{1+x^2}$	$\mathbb{R}$
$\text{arccot}\, x$	$-\frac{1}{1+x^2}$	$\mathbb{R}$
$\text{arcsec}\, x$	$\frac{1}{\vert x\vert \sqrt{x^2-1}}$	$\vert x\vert > 1$
$\text{arccsc}\, x$	$-\frac{1}{\vert x\vert \sqrt{x^2-1}}$	$\vert x\vert > 1$

Key Integrals

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

Key Identity

$$\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \quad \text{for all } x \in [-1, 1]$$

Common Student Questions

What's the difference between $\sin^{-1}(x)$ and $(\sin x)^{-1}$?

These are completely different!

$\sin^{-1}(x) = \arcsin(x)$ is the inverse function (the angle whose sine is $x$)
$(\sin x)^{-1} = \frac{1}{\sin x} = \csc x$ is the reciprocal (one divided by sine)

The notation is unfortunate, but standard. When in doubt, $\arcsin$ is unambiguous.

Why does the derivative of arcsin have a square root?

It comes from the Pythagorean identity! When we differentiate $\sin y = x$ implicitly: $$\cos y \cdot \frac{dy}{dx} = 1 \quad \Rightarrow \quad \frac{dy}{dx} = \frac{1}{\cos y}$$

Since $\sin y = x$ and $\sin^2 y + \cos^2 y = 1$, we get $\cos y = \sqrt{1-x^2}$.

When do I use $\arcsin$ vs. $\arctan$ for integrals?

Look at the form:

$\frac{1}{\sqrt{a^2 - x^2}}$ → arcsine (square root in denominator with minus sign)
$\frac{1}{a^2 + x^2}$ → arctangent (no square root, plus sign)

If there's an $x$ in the numerator, try u-substitution first. It might give you a logarithm instead.

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