Inverse Trigonometric Functions Revision Notes for AQA A-Level Further Maths

Inverse Trigonometric Functions

Introduction to differentiation of inverse trigonometric functions

When working with inverse trigonometric functions such as $\arcsin x$ , $\arccos x$ , and $\arctan x$ , you need a specific technique to find their derivatives. These functions are the inverses of the standard trigonometric functions, and differentiating them requires using the reciprocal rule for derivatives.

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The key relationship you must use is:

$\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}$

This formula tells you that the derivative of $y$ with respect to $x$ is the reciprocal of the derivative of $x$ with respect to $y$ . This is particularly useful when you can express $x$ in terms of $y$ more easily than $y$ in terms of $x$ .

Deriving the derivative of arcsin x

To understand how this works, let's derive the derivative of $y = \arcsin x$ step by step.

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Worked Example: Deriving the derivative of arcsin x

First, rearrange the equation to express $x$ in terms of $y$ : $x = \sin y$

Now differentiate both sides with respect to $y$ : $\frac{dx}{dy} = \cos y$

Using the reciprocal rule: $\frac{dy}{dx} = \frac{1}{\cos y}$

However, you need to express this in terms of $x$ , not $y$ . Use the Pythagorean identity $\cos^2 y + \sin^2 y = 1$ , which gives: $\cos y = \sqrt{1 - \sin^2 y}$

Since $x = \sin y$ , you can substitute: $\cos y = \sqrt{1 - x^2}$

Therefore: $\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}$

Key derivative formulas

The derivatives of the three main inverse trigonometric functions are:

$\frac{d(\arcsin x)}{dx} = \frac{1}{\sqrt{1-x^2}}$

$\frac{d(\arccos x)}{dx} = -\frac{1}{\sqrt{1-x^2}}$

$\frac{d(\arctan x)}{dx} = \frac{1}{1+x^2}$

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You must memorise these three formulas. Notice that the derivatives of $\arcsin x$ and $\arccos x$ differ only by a negative sign.

Worked example: Differentiating arctan 3x

Let's differentiate $y = \arctan 3x$ with respect to $x$ using the reciprocal rule method.

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Worked Example: Differentiating arctan 3x

Step 1: Express $x$ in terms of $y$ by rearranging: $3x = \tan y$ $x = \frac{1}{3}\tan y$

Step 2: Differentiate with respect to $y$ : $\frac{dx}{dy} = \frac{1}{3}\sec^2 y$

Step 3: Apply the reciprocal rule: $\frac{dy}{dx} = \frac{1}{\frac{1}{3}\sec^2 y} = \frac{3}{\sec^2 y}$

Step 4: Use the trigonometric identity $1 + \tan^2 y = \sec^2 y$ : $\frac{dy}{dx} = \frac{3}{1 + \tan^2 y}$

Step 5: Substitute back in terms of $x$ using $3x = \tan y$ : $\frac{dy}{dx} = \frac{3}{1 + (3x)^2} = \frac{3}{1 + 9x^2}$

Integration results from inverse trigonometric functions

By applying the fundamental theorem of calculus, you can reverse these derivative formulas to obtain important integration results. These are standard integrals you must know:

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

$\int \frac{1}{\sqrt{1-x^2}} \, dx = -\arccos x + c$

$\int \frac{1}{1+x^2} \, dx = \arctan x + c$

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These formulas are particularly useful when you encounter integrals involving square roots or quadratic denominators. The first two formulas give different results because the derivatives of $\arcsin x$ and $\arccos x$ differ by a sign.

Proving integration results using substitution

You can also derive these integration results using trigonometric substitution. This method provides deeper understanding and shows how the formulas connect to trigonometric identities.

Worked example: Proving the arcsin integration formula

Let's prove that $\int \frac{1}{\sqrt{1-x^2}} \, dx = \arcsin x + c$ using the substitution $x = \sin u$ .

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Worked Example: Proving the arcsin integration formula

Step 1: Make the substitution $x = \sin u$ : $\frac{dx}{du} = \cos u$

Therefore: $dx = \cos u \, du$

Step 2: Substitute into the integral: $\int \frac{1}{\sqrt{1-x^2}} \, dx = \int \frac{1}{\sqrt{1-\sin^2 u}} \cos u \, du$

Step 3: Apply the Pythagorean identity $1 - \sin^2 u = \cos^2 u$ : $= \int \frac{\cos u}{\sqrt{\cos^2 u}} \, du$

Step 4: Simplify (note that $\sqrt{\cos^2 u} = \cos u$ for the relevant range): $= \int 1 \, du$

Step 5: Integrate: $= u + c$

Step 6: Convert back to $x$ using $x = \sin u$ , which means $u = \arcsin x$ : $= \arcsin x + c$

Choosing appropriate substitutions

When you encounter more complex integrals, you need to choose suitable substitutions yourself. The exam will not always tell you which substitution to use. Here are the key patterns to recognise:

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Key substitution patterns:

For integrals involving $\sqrt{a^2 - x^2}$ , use the substitution $x = a\sin u$
For integrals involving $a^2 + x^2$ , use the substitution $x = a\tan u$

These substitutions work because they convert the expressions into simple trigonometric forms using the identities $1 - \sin^2 u = \cos^2 u$ and $1 + \tan^2 u = \sec^2 u$ .

Rearranging integrals before substitution

Sometimes you need to rearrange the integral first to match one of the standard forms. Look for opportunities to:

Factor out constants
Rewrite denominators in the form $a^2 + x^2$ or $a^2 - x^2$

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For example, $\int \frac{1}{1 + \frac{x^2}{4}} \, dx$ can be rewritten as $\int \frac{4}{4 + x^2} \, dx$ . You can then see this involves $a^2 + x^2$ where $a = 2$ , so use $x = 2\tan u$ .

Completing the square technique

Many exam questions require you to complete the square on the denominator before identifying the appropriate substitution. This is crucial when the denominator is a quadratic expression.

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Completing the square is essential when the denominator doesn't immediately match a standard form. This technique transforms complex quadratic expressions into recognizable patterns.

For example, to integrate $\int \frac{1}{\sqrt{8 + 2x - x^2}} \, dx$ :

Step 1: Complete the square:

$8 + 2x - x^2 = 9 - (x - 1)^2$

Step 2: Rewrite the integral:

$\int \frac{1}{\sqrt{9 - (x-1)^2}} \, dx$

Step 3: Now you can see this has the form $\sqrt{a^2 - x'^2}$ where $x' = x - 1$ and $a = 3$ . Use the substitution $x - 1 = 3\sin u$ .

Strategy for integrals requiring trigonometric substitution

Follow these steps systematically:

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Systematic approach:

Step 1: Rewrite the integrand so it involves either $\sqrt{a^2 - x^2}$ or $a^2 + x^2$

Step 2: Choose the correct substitution: either $x = a\sin u$ or $x = a\tan u$

Step 3: Use integration by substitution to work out the integral, then write the answer back in terms of $x$

Worked example: Integral requiring rearrangement

Let's work out $\int \frac{3}{\sqrt{1-9x^2}} \, dx$ .

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Worked Example: Integral requiring rearrangement

Step 1: Divide numerator and denominator by 3:

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

Step 2: Recognise this has the form $\sqrt{a^2 - x^2}$ where $a^2 = \frac{1}{9}$ , so $a = \frac{1}{3}$ . Let $x = \frac{1}{3}\sin u$ .

Step 3: Differentiate the substitution:

$\frac{dx}{du} = \frac{1}{3}\cos u$

Step 4: Substitute into the integral:

$\int \frac{1}{\sqrt{\frac{1}{9} - \left(\frac{1}{3}\sin u\right)^2}} \cdot \frac{1}{3}\cos u \, du = \frac{1}{3}\int \frac{\cos u}{\sqrt{\frac{1}{9} - \frac{1}{9}\sin^2 u}} \, du$

Step 5: Factor out $\frac{1}{9}$ from the square root:

$= \frac{1}{3}\int \frac{\cos u}{\frac{1}{3}\sqrt{1 - \sin^2 u}} \, du = \int \frac{\cos u}{\sqrt{\cos^2 u}} \, du$

Step 6: Simplify using $\sqrt{\cos^2 u} = \cos u$ :

$= \int 1 \, du = u + c$

Step 7: Convert back to $x$ using $x = \frac{1}{3}\sin u$ , so $u = \arcsin(3x)$ :

$= \arcsin(3x) + c$

Worked example: Integral with completing the square

Calculate $\int \frac{9}{x^2 + 6x + 25} \, dx$ .

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Worked Example: Integral with completing the square

Step 1: Complete the square on the denominator:

$x^2 + 6x + 25 = (x + 3)^2 + 16$

Step 2: Rewrite the integral:

$\int \frac{9}{(x+3)^2 + 16} \, dx$

Step 3: This has the form $a^2 + x'^2$ where $a = 4$ and $x' = x + 3$ .

Let $x + 3 = 4\tan u$ .

Step 4: Differentiate:

$\frac{dx}{du} = 4\sec^2 u$

Step 5: Substitute:

$\int \frac{9}{(4\tan u)^2 + 16} \cdot 4\sec^2 u \, du = \int \frac{36\sec^2 u}{16\tan^2 u + 16} \, du$

Step 6: Factor out 16 from the denominator:

$= \int \frac{36\sec^2 u}{16(\tan^2 u + 1)} \, du = \int \frac{36\sec^2 u}{16\sec^2 u} \, du$

Step 7: Simplify:

$= \int \frac{9}{4} \, du = \frac{9}{4}u + c$

Step 8: Convert back using $x + 3 = 4\tan u$ , so $u = \arctan\left(\frac{x+3}{4}\right)$ :

$= \frac{9}{4}\arctan\left(\frac{x+3}{4}\right) + c$

Exam tips

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Exam success strategies:

Always check if you need to complete the square before attempting substitution
Remember to convert your final answer back to terms of x
Watch out for constants that need factoring out before identifying the substitution
In the exam, if you're not told what substitution to use, look at the form of the integral to decide
The negative sign in the derivative of $\arccos x$ is easy to forget—take care with signs
When using $x = \sin u$ or $x = \tan u$ , remember to also substitute for $dx$

Remember!

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Key Points to Remember:

The derivative of $\arcsin x$ is $\frac{1}{\sqrt{1-x^2}}$ , of $\arccos x$ is $-\frac{1}{\sqrt{1-x^2}}$ , and of $\arctan x$ is $\frac{1}{1+x^2}$
Use the reciprocal rule $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}$ when differentiating inverse trigonometric functions
For integrals involving $\sqrt{a^2 - x^2}$ , use substitution $x = a\sin u$ ; for $a^2 + x^2$ , use $x = a\tan u$
Always complete the square first if the denominator is a quadratic expression that doesn't immediately match a standard form
Remember to convert your final answer back to terms of $x$ after using substitution in integration problems

Inverse Trigonometric Functions (AQA A-Level Further Maths): Revision Notes