Ultimate Guide to Trigonometric Integration and t-formulas

Introduction

In HSC Mathematics Advanced and Extension 1, you learned how to integrate basic trigonometric functions and squares like $\sin^2 x$ and $\cos^2 x$ . In HSC Mathematics Extension 2, the complexity ramps up. How do you integrate a product of two different trigonometric functions with different periods, like $\sin(3x)\cos(5x)$ ? Furthermore, how do you handle fractions containing trigonometric terms in the denominator? This guide covers the essential identities that turn impossible trigonometric integrals into simple sums, and introduces the universal substitution known as the $t$ -formulas.

Executive Summary

This guide breaks down two core trigonometric techniques:

Products to Sums: Deriving and applying the identities that convert products like $\sin(A)\cos(B)$ into sums like $\frac{1}{2}[\sin(A+B) + \sin(A-B)]$ to make integration possible.
The $t$ -formulas: Defining $t = \tan(\frac{x}{2})$ to express $\sin x$ , $\cos x$ , and $\tan x$ entirely in terms of the algebraic variable $t$ .
Solving Equations: Using $t$ -formulas to solve trigonometric equations over restricted domains.

What is this about?

Integration is a linear operator. This means it is very easy to integrate a sum of functions (you just integrate each piece separately), but it is generally very difficult to integrate a product of functions. When faced with $\int \sin(3x)\cos(5x) \, dx$ , you cannot just integrate the sine and the cosine separately.

To solve this, we use Products to Sums identities to rewrite the product as a sum of two separate trigonometric terms, which can then be integrated easily.

Additionally, equations like $3\sin x + 4\cos x = 5$ can be solved using auxiliary angles (from Ext 1), but the $t$ -formulas provide a purely algebraic alternative. By substituting $t = \tan(\frac{x}{2})$ , every trigonometric term is converted into a polynomial or rational fraction in $t$ , turning a trigonometry problem into a standard algebra problem.

Main Content

1. Products to Sums Identities

The syllabus requires you to be able to derive these formulas from the standard compound angle formulas learned in Extension 1.

Derivation Example: We know the compound angle formulas for sine:

$\sin(A + B) = \sin A \cos B + \cos A \sin B$
$\sin(A - B) = \sin A \cos B - \cos A \sin B$

If we add equation (1) and equation (2) together, the $\cos A \sin B$ terms cancel out: $\sin(A + B) + \sin(A - B) = 2\sin A \cos B$

Rearranging this gives us our first product-to-sum identity: $\sin A \cos B = \frac{1}{2} [\sin(A+B) + \sin(A-B)]$

By adding or subtracting the compound angle formulas for cosine, we can derive the other two essential identities: $\cos A \cos B = \frac{1}{2} [\cos(A+B) + \cos(A-B)]$ $\sin A \sin B = \frac{1}{2} [\cos(A-B) - \cos(A+B)]$ (Note the order in the last one! It is $A-B$ minus $A+B$ to avoid negative signs).

2. Integrating Trigonometric Products

Once you have converted the product into a sum, integration is straightforward.

Example: Evaluate $\int \sin(3x) \cos(2x) \, dx$ . Here, $A = 3x$ and $B = 2x$ . $\sin(3x) \cos(2x) = \frac{1}{2} [\sin(3x+2x) + \sin(3x-2x)]$ $= \frac{1}{2} [\sin(5x) + \sin(x)]$

Now, integrate the sum: $\int \frac{1}{2} [\sin(5x) + \sin(x)] \, dx$ $= \frac{1}{2} \left[ -\frac{1}{5}\cos(5x) - \cos(x) \right] + C$ $= -\frac{1}{10}\cos(5x) - \frac{1}{2}\cos(x) + C$

3. The $t$ -formulas

The $t$ -formulas are based on the substitution $t = \tan\left(\frac{x}{2}\right)$ . Using double angle formulas and right-angled triangle geometry, we can derive expressions for $\sin x$ , $\cos x$ , and $\tan x$ entirely in terms of $t$ :

$\sin x = \frac{2t}{1 + t^2}$ $\cos x = \frac{1 - t^2}{1 + t^2}$ $\tan x = \frac{2t}{1 - t^2}$

Syllabus Note: You must be able to derive these. Start with $\sin x = \sin(2 \times \frac{x}{2}) = 2\sin(\frac{x}{2})\cos(\frac{x}{2})$ , then divide by $\cos^2(\frac{x}{2}) + \sin^2(\frac{x}{2})$ (which equals 1), and divide top and bottom by $\cos^2(\frac{x}{2})$ to introduce $\tan(\frac{x}{2})$ .

4. Solving Equations with $t$ -formulas

When solving an equation like $a\sin x + b\cos x = c$ , you can substitute the $t$ -formulas to create a rational algebraic equation.

Warning: The Domain Restriction When you use the substitution $t = \tan(\frac{x}{2})$ , you are assuming that $\tan(\frac{x}{2})$ is defined. However, the tangent function has asymptotes. $\tan(\frac{x}{2})$ is undefined when $\frac{x}{2} = \frac{\pi}{2}$ (i.e., when $x = \pi, 3\pi, -\pi$ , etc.). If you use the $t$ -formulas, you must manually check if $x = \pi$ is a valid solution to the original equation, because the $t$ -formula method will completely "miss" this solution.

Simple Worked Example

Question: Solve the equation $3\cos x - 4\sin x = 3$ for $0 \le x \le 2\pi$ using the $t$ -formulas.

Solution: Step 1: Substitute the $t$ -formulas. Let $t = \tan\left(\frac{x}{2}\right)$ . Substitute $\cos x = \frac{1-t^2}{1+t^2}$ and $\sin x = \frac{2t}{1+t^2}$ into the equation: $3\left(\frac{1-t^2}{1+t^2}\right) - 4\left(\frac{2t}{1+t^2}\right) = 3$

Step 2: Solve the algebraic equation. Multiply the entire equation by $(1+t^2)$ to clear denominators: $3(1-t^2) - 4(2t) = 3(1+t^2)$ $3 - 3t^2 - 8t = 3 + 3t^2$ Move all terms to one side: $0 = 6t^2 + 8t$ $0 = 2t(3t + 4)$

So, $t = 0$ or $t = -\frac{4}{3}$ .

Step 3: Convert $t$ back to $x$ . Recall that $t = \tan\left(\frac{x}{2}\right)$ . If $t = 0$ : $\tan\left(\frac{x}{2}\right) = 0$ $\frac{x}{2} = 0, \pi, 2\pi$ $x = 0, 2\pi, 4\pi$ Within the domain $0 \le x \le 2\pi$ , the solutions are $x = 0$ and $x = 2\pi$ .

If $t = -\frac{4}{3}$ : $\tan\left(\frac{x}{2}\right) = -\frac{4}{3}$ Since tangent is negative, $\frac{x}{2}$ is in the 2nd quadrant. $\frac{x}{2} = \pi - \tan^{-1}\left(\frac{4}{3}\right) \approx 2.214$ $x = 2 \times 2.214 \approx 4.43$ radians.

Step 4: Check the undefined case. Is $x = \pi$ a solution? LHS = $3\cos(\pi) - 4\sin(\pi) = 3(-1) - 4(0) = -3$ . RHS = $3$ . Since $-3 \neq 3$ , $x = \pi$ is not a solution.

Final Answer: $x = 0, 2\pi$ , and $x \approx 4.43$ .

mini-FAQ page

Q: Should I use auxiliary angles ( $R\cos(x-\alpha)$ ) or $t$ -formulas to solve $a\sin x + b\cos x = c$ ? A: Both are valid. The auxiliary angle method is usually faster and less prone to algebraic errors, and it doesn't require checking the $x=\pi$ case. However, if the question specifically says "using the $t$ -formulas," you must use this algebraic approach to get the marks.

Q: Do I need to memorize the product-to-sum formulas? A: They are usually provided on the HSC Reference Sheet, but the syllabus requires you to be able to derive them. Understanding the derivation makes them much easier to remember and use under exam pressure.

Common mistakes to avoid

Forgetting the $\frac{1}{2}$ in product-to-sums: A very common mistake is writing $\sin A \cos B = \sin(A+B) + \sin(A-B)$ . You must remember the factor of $\frac{1}{2}$ out the front.
Forgetting to manually check $x=\pi$ : If you use $t$ -formulas and the true solution includes $x=\pi$ , the algebra will simply not yield it. You will lose marks if you do not show that you manually tested the boundary condition.

Practice on Vu's Maths Hub

Trigonometric integration requires a sharp memory for identities.

Practice deriving identities and solving $t$ -formula equations in the HSC Integration Booklet.
Review your foundational trigonometry in the HSC Trigonometry Booklet.

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