Calculus with Inverse Trig Functions

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

Starting point: what you already know

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

Finding the derivatives

Let $y = sin^{- 1} x$ , so $x = sin y$ with $- \frac{π}{2} \leq y \leq \frac{π}{2}$ .

$\frac{d x}{d y} = cos y = 1 - sin^{2} y = 1 - x^{2}$

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

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

The same idea gives the other two.

Integrals that give inverse trig

Reverse the derivatives:

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 12Harder forms

Take out a constant to match the standard form: $\int \frac{d x}{9 - 4 x ^{2}} = \int \frac{d x}{2 \frac{9}{4} - x ^{2}} = \frac{1}{2} sin^{- 1} \frac{2 x}{3} + C$ And $\int_{0}^{1} \frac{d x}{1 + x ^{2}} = tan^{- 1} 1 = \frac{π}{4}$ , a surprising way to meet $π$ .

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

A substitution. $\int_{0}^{1} \frac{x}{1 + x ^{4}} d x$ : let $u = x^{2}$ , so $d u = 2 x d x$ and $x^{4} = u^{2}$ . It becomes $\frac{1}{2} \int_{0}^{1} \frac{d u}{1 + u ^{2}} = \frac{π}{8}$ .

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

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

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

Differentiate, hence integrate. $\frac{d}{d x} (x sin^{- 1} x + 1 - x^{2}) = sin^{- 1} x$ , so $\int sin^{- 1} x d x = x sin^{- 1} x + 1 - x^{2} + C$ .

Quick summary

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

Quiz

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Try 3 questions now

$\frac{d}{d x} sin^{- 1} x$ is:
1. A $\frac{1}{1 - x ^{2}}$
2. B $\frac{1}{1 + x ^{2}}$
3. C $cos^{- 1} x$
4. D $\frac{- 1}{1 - x ^{2}}$
Show answer
A: $\frac{1}{1 - x ^{2}}$
From $x = sin y$ and $\frac{d x}{d y} = cos y$ .
$\frac{d}{d x} tan^{- 1} x$ is:
1. A $sec^{2} x$
2. B $\frac{1}{1 + x ^{2}}$
3. C $\frac{1}{1 + x ^{2}}$
4. D $\frac{1}{cos ^{2} x}$
Show answer
B: $\frac{1}{1 + x ^{2}}$
The derivative has no trig in it.
$\frac{d}{d x} cos^{- 1} x$ is:
1. A $\frac{1}{1 - x ^{2}}$
2. B $\frac{- 1}{1 - x ^{2}}$
3. C $- sin^{- 1} x$
4. D $\frac{- 1}{1 + x ^{2}}$
Show answer
B: $\frac{- 1}{1 - x ^{2}}$
Opposite sign to $sin^{- 1} x$ .