Calculus of Inverse Trigonometrical Functions (FP2.7)
Welcome to a chapter where we unlock the calculus secrets of one of the most useful groups of functions: the Inverse Trigonometrical Functions (also known as Arc Functions).
You’ve mastered differentiating and integrating sine, cosine, and tangent. Now, we turn our attention to the functions that "undo" them: \(\sin^{-1} x\), \(\cos^{-1} x\), and \(\tan^{-1} x\). These concepts are absolutely crucial for solving specific, challenging integration problems, particularly when dealing with fractional expressions involving square roots or sums of squares.
Quick Review: What are Inverse Trig Functions?
Think of an inverse trig function as the mathematical way of asking a question about an angle:
If \(\sin(\theta) = x\), then the inverse function asks: What is \(\theta\)?
We write the answer as \(\theta = \sin^{-1} x\) (or \(\theta = \arcsin x\)).
- Notation Check: In Further Maths, we mainly use the notation \(\sin^{-1} x\), \(\cos^{-1} x\), and \(\tan^{-1} x\).
- Domain & Range: Remember that inverse functions are restricted to principal value ranges (e.g., \(\sin^{-1} x\) is usually restricted to \([-\frac{\pi}{2}, \frac{\pi}{2}]\)) to ensure they remain functions.
1. Differentiation of Inverse Trig Functions
Don't worry about deriving these formulas in the exam—they are provided in your formulae booklet! Your job is to know them and apply them correctly, especially when the argument of the function is more complex than just \(x\).
The Key Derivatives (From the Formulae Booklet)
These are the fundamental results you must be familiar with:
(i) Derivative of \(y = \sin^{-1} x\)
If \(y = \sin^{-1} x\), then:
(ii) Derivative of \(y = \cos^{-1} x\)
If \(y = \cos^{-1} x\), then:
Did You Spot the Trick?
Notice how the derivative of \(\cos^{-1} x\) is exactly the same as the derivative of \(\sin^{-1} x\), but with a negative sign! This is a great memory aid. If you know one, you know the other.
(iii) Derivative of \(y = \tan^{-1} x\)
If \(y = \tan^{-1} x\), then:
Quick Review: Key Derivatives
- \(\sin^{-1} x \rightarrow\) positive fraction with root.
- \(\cos^{-1} x \rightarrow\) negative fraction with root.
- \(\tan^{-1} x \rightarrow\) positive fraction without root.
Applying the Chain Rule
When the argument is a function of \(x\), say \(y = \sin^{-1} (f(x))\), you must use the chain rule. This means multiplying the standard derivative by the derivative of the inner function, \(f'(x)\).
Example: Find the derivative of \(y = \tan^{-1} (2x)\).
Let \(u = 2x\). Then \(\frac{du}{dx} = 2\).
Using the chain rule, \(\frac{dy}{dx} = \frac{1}{1 + u^2} \cdot \frac{du}{dx}\).
\[ \frac{dy}{dx} = \frac{1}{1 + (2x)^2} \cdot 2 = \frac{2}{1 + 4x^2} \]
Key Takeaway for Differentiation: Know the three standard derivatives (they are in the booklet!). Remember to use the chain rule when the argument is not simply \(x\).
2. Integration Leading to Inverse Trig Functions
Integration is simply the reverse of differentiation. Therefore, if we integrate the standard derivatives, we return to the inverse trig functions. These integrals are called Standard Integrals and are essential tools for FP2.
The standard forms involving a constant \(a\) are also provided in your formulae booklet. These are the two forms you must be able to recognise and use:
The Standard Integrals
(i) Integral leading to \(\tan^{-1}\) (The Sum of Squares)
This integral arises from the derivative of \(\tan^{-1} x\), generalised to include a constant, \(a\).
Analogy: The Perfect Denominator
If you see a denominator that looks like a number squared plus a variable squared (\(a^2 + x^2\)), think \(\tan^{-1}\). The result always includes a factor of \(\frac{1}{a}\) outside and \(\frac{x}{a}\) inside the inverse tangent function.
Common Mistake to Avoid: Forgetting the \(\frac{1}{a}\) factor at the front of the result!
(ii) Integral leading to \(\sin^{-1}\) (The Difference of Squares under a Root)
This integral arises from the derivative of \(\sin^{-1} x\), generalised to include a constant, \(a\).
Accessibility Note: The Crucial Order
For the inverse sine integral, the form must be \(\sqrt{\mathbf{a^2 - x^2}}\). The constant squared MUST come first under the root sign.
Did you know? If the order were reversed, \(\sqrt{x^2 - a^2}\), the answer would involve inverse hyperbolic cosine (\(\cosh^{-1}\)), which is covered in FP2.9. Since the syllabus for FP2.7 only includes the inverse sine form, focus only on \(\sqrt{a^2 - x^2}\).
Key Takeaway for Integration: The constant \(a\) is critical. You must identify what \(a^2\) is in the denominator (and hence what \(a\) is) to correctly substitute into the final formula.
3. Strategy: Manipulating Integrals to Fit the Standard Forms
Often, the integral you are faced with won't look exactly like the standard forms, but you can manipulate them using two key techniques: Completing the Square and Substitution.
3.1 Completing the Square
If the denominator is a quadratic expression, you may need to complete the square to achieve the standard form, \(\int \frac{1}{(A)^2 + (f(x))^2} \, dx\) or \(\int \frac{1}{\sqrt{(A)^2 - (f(x))^2}} \, dx\).
Step-by-Step Example (The \(\tan^{-1}\) case):
Consider \(\int \frac{1}{x^2 + 4x + 5} \, dx\).
- Complete the Square: \(x^2 + 4x + 5 = (x + 2)^2 - 4 + 5 = (x + 2)^2 + 1\).
- Rewrite the Integral: \[ \int \frac{1}{(x + 2)^2 + 1} \, dx \]
- Identify Parameters: Compare this to \(\int \frac{1}{u^2 + a^2} \, du\).
Here, \(u = x + 2\) (so \(du = dx\)) and \(a^2 = 1\), meaning \(a = 1\). - Apply the Formula: The integral is \(\frac{1}{a} \tan^{-1} \left( \frac{u}{a} \right) + C\).
\[ \frac{1}{1} \tan^{-1} \left( \frac{x + 2}{1} \right) + C = \tan^{-1} (x + 2) + C \]
3.2 Using Substitution (The Chain Rule in Reverse)
If the expression involving \(x\) inside the inverse trig function is complex, or if the numerator contains the derivative of the denominator's argument, you will need substitution.
If you see \(\int \frac{f'(x)}{\sqrt{a^2 - (f(x))^2}} \, dx\), you should use the substitution \(u = f(x)\).
Example: Evaluate \(\int \frac{2x}{\sqrt{4 - x^4}} \, dx\)
This looks complicated! Let's try to fit it to the \(\sin^{-1}\) form, \(\int \frac{1}{\sqrt{a^2 - u^2}} \, du\).
- Choose Substitution: Notice that the argument under the root is \(x^4 = (x^2)^2\). Let \(u = x^2\).
- Find \(du\): \(\frac{du}{dx} = 2x\), so \(du = 2x \, dx\).
- Rewrite the Integral: The numerator \(2x \, dx\) becomes \(du\), and the denominator becomes \(\sqrt{4 - u^2}\).
\[ \int \frac{du}{\sqrt{4 - u^2}} \] - Identify Parameters: Compare this to \(\int \frac{1}{\sqrt{a^2 - u^2}} \, du\).
Here, \(a^2 = 4\), so \(a = 2\). - Apply the Formula: The integral is \(\sin^{-1} \left( \frac{u}{a} \right) + C\).
\[ \sin^{-1} \left( \frac{u}{2} \right) + C \] - Return to \(x\): Substitute \(u = x^2\) back in.
\[ \sin^{-1} \left( \frac{x^2}{2} \right) + C \]
Encouragement: Don't worry if substitution seems challenging; the key is recognising the pattern. If you see \(\frac{f'(x)}{\text{function based on } f(x)}\), substitution is usually the way forward!
Quick Review: Inverse Trig Calculus Checklist
Differentiation (\(\frac{d}{dx}\)):
1. \(\sin^{-1} x\) gives \(\frac{1}{\sqrt{1 - x^2}}\).
2. \(\cos^{-1} x\) gives \(-\frac{1}{\sqrt{1 - x^2}}\).
3. \(\tan^{-1} x\) gives \(\frac{1}{1 + x^2}\).
*Always use the Chain Rule if the argument is complex.*
Integration (\(\int\)):
1. \(\int \frac{1}{a^2 + x^2} \, dx\) (Sum of squares) leads to \(\frac{1}{a} \tan^{-1} (\frac{x}{a}) + C\).
2. \(\int \frac{1}{\sqrt{a^2 - x^2}} \, dx\) (Constant minus variable squared under root) leads to \(\sin^{-1} (\frac{x}{a}) + C\).
*Look out for Completing the Square or Substitution to transform the integral into a standard form.*