# Calculus with Inverse Trig Functions

> Differentiate sin⁻¹ x, cos⁻¹ x and tan⁻¹ x, and use them to integrate 1/√(a² − x²) and 1/(a² + x²).

From HSC Maths Extension 1 on WhizzNotes, free study notes for NSW students: https://whizznotes.com/mathematics-extension-1/calculus-of-inverse-trig/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: The lighthouse beam
A lighthouse stands 1 km from a straight shore. Its beam sweeps along the beach. When the bright spot is $x$ km along the shore, the beam's angle is $\theta = \tan^{-1} x$.

The keeper wants to know how fast the angle changes as the spot moves. That needs the derivative of $\tan^{-1} x$. The surprising result is a simple fraction, $\dfrac{1}{1 + x^2}$, with no trig in it at all. Running the result backwards also lets you integrate functions that looked impossible before.

**In short:** The inverse trig functions have simple derivatives: $\dfrac{d}{dx}\sin^{-1} x = \dfrac{1}{\sqrt{1 - x^2}}$ and $\dfrac{d}{dx}\tan^{-1} x = \dfrac{1}{1 + x^2}$. Reading these backwards gives integrals such as $\int \dfrac{dx}{a^2 + x^2} = \dfrac{1}{a}\tan^{-1}\dfrac{x}{a} + C$.

## Starting point: what you already know

- **Inverse trig functions** and their domains and ranges (Trigonometric Functions unit).
- From Mathematics Advanced: the chain rule $\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}$, and $\dfrac{d}{dx}\sin x = \cos x$, $\dfrac{d}{dx}\tan x = \sec^2 x$.
- $\dfrac{dy}{dx} = \dfrac{1}{dx/dy}$, and the identities $\sin^2\theta + \cos^2\theta = 1$ and $1 + \tan^2\theta = \sec^2\theta$.
- Integration is the reverse of differentiation: $\displaystyle\int f'(x)\,dx = f(x) + C$.

## Finding the derivatives

Let $y = \sin^{-1} x$, so $x = \sin y$ with $-\dfrac{\pi}{2} \le y \le \dfrac{\pi}{2}$.

$$\frac{dx}{dy} = \cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$$

We took the positive root because $\cos y \ge 0$ in that range. Flip it over:

$$\frac{d}{dx}\sin^{-1} x = \frac{1}{\sqrt{1 - x^2}}$$

The same idea gives the other two.

### Key idea: Derivatives
$$\frac{d}{dx}\sin^{-1}\frac{x}{a} = \frac{1}{\sqrt{a^2 - x^2}} \qquad \frac{d}{dx}\cos^{-1}\frac{x}{a} = \frac{-1}{\sqrt{a^2 - x^2}} \qquad \frac{d}{dx}\tan^{-1}\frac{x}{a} = \frac{a}{a^2 + x^2}$$
With $a = 1$ these are the basic forms.

### Worked example: Chain rule
$\dfrac{d}{dx}\tan^{-1}(3x) = \dfrac{1}{1 + (3x)^2} \times 3 = \dfrac{3}{1 + 9x^2}$

### Worked example: Product rule
$\dfrac{d}{dx}\left(x\sin^{-1} x\right) = \sin^{-1} x + \dfrac{x}{\sqrt{1 - x^2}}$

### Exam tip: Why sin⁻¹ and cos⁻¹ have opposite derivatives
$\sin^{-1} x + \cos^{-1} x = \dfrac{\pi}{2}$ is a constant, so their derivatives must add to zero.

## Integrals that give inverse trig

Reverse the derivatives:

### Key idea: Standard integrals
$$\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\frac{x}{a} + C \qquad \int \frac{dx}{a^2 + x^2} = \frac{1}{a}\tan^{-1}\frac{x}{a} + C$$

### Worked example: A definite integral
$$\int_0^{2} \frac{dx}{4 + x^2} = \left[\frac{1}{2}\tan^{-1}\frac{x}{2}\right]_0^2 = \frac{1}{2}\cdot\frac{\pi}{4} - 0 = \frac{\pi}{8}$$

### Worked example: Spotting the form
$$\int_0^{3/2} \frac{dx}{\sqrt{9 - x^2}} = \left[\sin^{-1}\frac{x}{3}\right]_0^{3/2} = \sin^{-1}\frac{1}{2} = \frac{\pi}{6}$$

### Watch out: Read the shape carefully
$\dfrac{1}{\sqrt{a^2 - x^2}}$ (square root, minus) gives $\sin^{-1}$. $\dfrac{1}{a^2 + x^2}$ (no root, plus) gives $\tan^{-1}$ with an extra $\dfrac{1}{a}$ in front.

### Calculator tip: Exact values in radian mode
In radian mode, $\tan^{-1}1$ gives 0.785. Many calculators show it exactly as $\frac{1}{4}\pi$ (press `S⇔D` to switch). Give exact answers such as $\frac{\pi}{4}$ unless the question asks for a decimal.

**Mind map**

```text
Calculus with inverse trig
  Derivatives
    sin⁻¹ x: 1/√(1 - x²)
    cos⁻¹ x: -1/√(1 - x²)
    tan⁻¹ x: 1/(1 + x²)
  With a
    sin⁻¹(x/a): 1/√(a² - x²)
    tan⁻¹(x/a): a/(a² + x²)
  Integrals
    1/√(a² - x²) gives sin⁻¹(x/a)
    1/(a² + x²) gives (1/a) tan⁻¹(x/a)
  Skills
    Chain rule
    Spot the form
```

### Year 11: Back to the lighthouse
$\theta = \tan^{-1} x$, so $\dfrac{d\theta}{dx} = \dfrac{1}{1 + x^2}$. Near the lighthouse, at $x = 0$, the angle changes at 1 radian per km. At $x = 3$ km it changes at only $\dfrac{1}{10}$ radian per km. The spot races along distant parts of the beach for tiny turns of the lamp.

### Year 12: Harder forms
Take out a constant to match the standard form:
$$\int \frac{dx}{\sqrt{9 - 4x^2}} = \int \frac{dx}{2\sqrt{\tfrac{9}{4} - x^2}} = \frac{1}{2}\sin^{-1}\frac{2x}{3} + C$$
And $\displaystyle\int_0^1 \frac{dx}{1 + x^2} = \tan^{-1} 1 = \frac{\pi}{4}$, a surprising way to meet $\pi$.

**Reverse chain rule.** If the derivative of the inside is present: $\displaystyle\int \frac{e^{\tan^{-1}x}}{1 + x^2}\,dx = e^{\tan^{-1}x} + C$, because $\frac{1}{1 + x^2}$ is the derivative of $\tan^{-1}x$.

**A substitution.** $\displaystyle\int_0^1 \frac{x}{1 + x^4}\,dx$: let $u = x^2$, so $du = 2x\,dx$ and $x^4 = u^2$. It becomes $\frac{1}{2}\displaystyle\int_0^1 \frac{du}{1 + u^2} = \frac{\pi}{8}$.

**Complete the square.** $\displaystyle\int \frac{dx}{x^2 + 6x + 10} = \int \frac{dx}{(x + 3)^2 + 1} = \tan^{-1}(x + 3) + C$.

**Showing an expression is constant.** $\dfrac{d}{dx}\tan^{-1}\dfrac{1}{x} = \dfrac{1}{1 + 1/x^2} \times \left(-\dfrac{1}{x^2}\right) = -\dfrac{1}{1 + x^2}$, so $\tan^{-1}x + \tan^{-1}\frac{1}{x}$ has derivative 0 and is constant ($\frac{\pi}{2}$ for $x > 0$).

**Tangents.** The tangent to $y = \tan^{-1}x$ at $x = 1$ passes through $\left(1, \frac{\pi}{4}\right)$ with gradient $\frac{1}{2}$.

**Differentiate, hence integrate.** $\dfrac{d}{dx}\left(x\sin^{-1}x + \sqrt{1 - x^2}\right) = \sin^{-1}x$, so $\displaystyle\int \sin^{-1}x\,dx = x\sin^{-1}x + \sqrt{1 - x^2} + C$.

## Quick summary

- $\dfrac{d}{dx}\sin^{-1} x = \dfrac{1}{\sqrt{1 - x^2}}$, $\dfrac{d}{dx}\cos^{-1} x = \dfrac{-1}{\sqrt{1 - x^2}}$, $\dfrac{d}{dx}\tan^{-1} x = \dfrac{1}{1 + x^2}$.
- Use the chain rule for $\sin^{-1}(f(x))$ and similar.
- $\displaystyle\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\frac{x}{a} + C$ and $\displaystyle\int \frac{dx}{a^2 + x^2} = \frac{1}{a}\tan^{-1}\frac{x}{a} + C$.

## Quiz

Try these questions from the chapter quiz.

1. $\dfrac{d}{dx}\sin^{-1} x$ is:
   - A. $\dfrac{1}{\sqrt{1 - x^2}}$
   - B. $\dfrac{1}{1 + x^2}$
   - C. $\cos^{-1} x$
   - D. $\dfrac{-1}{\sqrt{1 - x^2}}$
   - Answer: A. $\dfrac{1}{\sqrt{1 - x^2}}$. From $x = \sin y$ and $\dfrac{dx}{dy} = \cos y$.

2. $\dfrac{d}{dx}\tan^{-1} x$ is:
   - A. $\sec^2 x$
   - B. $\dfrac{1}{1 + x^2}$
   - C. $\dfrac{1}{\sqrt{1 + x^2}}$
   - D. $\dfrac{1}{\cos^2 x}$
   - Answer: B. $\dfrac{1}{1 + x^2}$. The derivative has no trig in it.

3. $\dfrac{d}{dx}\cos^{-1} x$ is:
   - A. $\dfrac{1}{\sqrt{1 - x^2}}$
   - B. $\dfrac{-1}{\sqrt{1 - x^2}}$
   - C. $-\sin^{-1} x$
   - D. $\dfrac{-1}{1 + x^2}$
   - Answer: B. $\dfrac{-1}{\sqrt{1 - x^2}}$. Opposite sign to $\sin^{-1} x$.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/mathematics-extension-1/calculus-of-inverse-trig/#quiz