# Integrating with Trig Identities

> Integrate sin² x, cos² x, sin x cos x and tan² x by first rewriting them with the double angle formulas.

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

### Story time: The flickering light
Mains power makes a light bulb's brightness rise and fall about 100 times a second. The power at any moment is shaped like $\sin^2 t$. An engineer wants the average power, which needs the area under $\sin^2 t$.

There is no rule for integrating $\sin^2 t$ straight away. But a trig identity rewrites it as something with no squares in it. Then it takes one line. The answer also shows why the average of $\sin^2 t$ is exactly one half.

**In short:** Some trig integrals need an identity first. Use $\sin^2 x = \dfrac{1}{2}(1 - \cos 2x)$ and $\cos^2 x = \dfrac{1}{2}(1 + \cos 2x)$ to integrate squares, and $\tan^2 x = \sec^2 x - 1$ for $\tan^2 x$.

## Starting point: what you already know

- **Double angle formulas** (Compound Angles and Double Angles): $\cos 2x = 1 - 2\sin^2 x = 2\cos^2 x - 1$ and $\sin 2x = 2\sin x\cos x$.
- **Pythagorean identities** (Reciprocal Trigonometric Functions): $\sin^2 x + \cos^2 x = 1$ and $1 + \tan^2 x = \sec^2 x$.
- **Integrals of trig functions** from Mathematics Advanced:
  $\displaystyle\int \cos ax\,dx = \frac{1}{a}\sin ax + C$, $\displaystyle\int \sin ax\,dx = -\frac{1}{a}\cos ax + C$, $\displaystyle\int \sec^2 ax\,dx = \frac{1}{a}\tan ax + C$.
- **Radians:** calculus with trig always uses radians.

## Rewrite the squares

Rearrange the double angle formulas for $\cos 2x$:

### Key idea: Squares to double angles
$$\sin^2 x = \frac{1}{2}(1 - \cos 2x) \qquad \cos^2 x = \frac{1}{2}(1 + \cos 2x)$$

Now each one is easy to integrate:

$$\int \sin^2 x\,dx = \frac{x}{2} - \frac{\sin 2x}{4} + C \qquad \int \cos^2 x\,dx = \frac{x}{2} + \frac{\sin 2x}{4} + C$$

### Worked example: A definite integral
$$\int_0^{\pi/2} \cos^2\theta\,d\theta = \left[\frac{\theta}{2} + \frac{\sin 2\theta}{4}\right]_0^{\pi/2} = \frac{\pi}{4} + \frac{\sin\pi}{4} - 0 = \frac{\pi}{4}$$

This is the quarter circle area from Integration by Substitution.

*Diagram: Graph of y equals sine squared x from 0 to 2 pi. It rises and falls between 0 and 1, and is balanced around the dashed line y equals one half.*
*y = sin² x swings between 0 and 1. The humps above the dashed line exactly fill the dips below it, so the average is one half.*

### Calculator tip: Radian mode for trig integrals
Calculus with trig functions always uses radians. Switch to radian mode before checking an answer such as $\displaystyle\int_0^{\pi/4}\tan^2 x\,dx = 1 - \frac{\pi}{4} \approx 0.215$ with the integral key, if your calculator has one.

## Other products and powers

### Worked example: sin x cos x
$\sin x\cos x = \dfrac{1}{2}\sin 2x$, so
$$\int \sin x\cos x\,dx = -\frac{1}{4}\cos 2x + C$$

### Worked example: A different angle inside
$\sin^2 3x = \dfrac{1}{2}(1 - \cos 6x)$. Double the angle inside, not outside.
$$\int \sin^2 3x\,dx = \frac{x}{2} - \frac{\sin 6x}{12} + C$$

### Worked example: tan² x
$\tan^2 x = \sec^2 x - 1$, so
$$\int \tan^2 x\,dx = \tan x - x + C$$

### Watch out: sin² x is not sin(x²)
$\sin^2 x$ means $(\sin x)^2$. The rule $\displaystyle\int \sin^2 x\,dx = -\frac{\cos^3 x}{3}$ is wrong. Always use the identity.

**Mind map**

```text
Integrating with trig identities
  Squares
    sin² x = ½(1 - cos 2x)
    cos² x = ½(1 + cos 2x)
  Products
    sin x cos x = ½ sin 2x
  tan² x
    sec² x - 1
    Gives tan x - x
  Inside angle
    sin² 3x uses cos 6x
  Uses
    Average value
    Volumes of solids
```

### Year 11: The average of sin² t
Over one full turn, $\displaystyle\int_0^{2\pi} \sin^2 t\,dt = \left[\frac{t}{2} - \frac{\sin 2t}{4}\right]_0^{2\pi} = \pi$.

The average value is the area divided by the width: $\dfrac{\pi}{2\pi} = \dfrac{1}{2}$. That is why the average power of a mains light is half its peak power.

### Year 11: A definite tan² integral
$$\int_0^{\pi/4} \tan^2 x\,dx = \big[\tan x - x\big]_0^{\pi/4} = 1 - \frac{\pi}{4}$$

**Products to sums.** For a product of different angles, use $2\sin A\cos B = \sin(A + B) + \sin(A - B)$: $\displaystyle\int 2\sin 5x\cos x\,dx = \int (\sin 6x + \sin 4x)\,dx = -\frac{\cos 6x}{6} - \frac{\cos 4x}{4} + C$.

**Half angles.** $\cos^2\frac{x}{2} = \frac{1}{2}(1 + \cos x)$, because doubling $\frac{x}{2}$ gives $x$.

**Expand first.** $(\sin x + \cos x)^2 = 1 + \sin 2x$, which integrates to $x - \frac{1}{2}\cos 2x + C$.

**Fourth powers.** Use the identity twice: $\sin^4 x = \frac{1}{4}(1 - \cos 2x)^2 = \frac{1}{4}\left(1 - 2\cos 2x + \frac{1 + \cos 4x}{2}\right)$. Over $0$ to $\frac{\pi}{2}$ the cosine terms integrate to 0, leaving $\frac{3}{8} \times \frac{\pi}{2} = \frac{3\pi}{16}$.

**A half-angle trick.** $1 + \cos x = 2\cos^2\frac{x}{2}$, so $\displaystyle\int \frac{dx}{1 + \cos x} = \frac{1}{2}\int \sec^2\frac{x}{2}\,dx = \tan\frac{x}{2} + C$.

### Year 12: Odd powers: save one factor
For $\displaystyle\int \cos^3 x\,dx$, keep one $\cos x$ and change the rest with $\cos^2 x = 1 - \sin^2 x$:
$$\int (1 - \sin^2 x)\cos x\,dx$$
Now let $u = \sin x$: $\displaystyle\int (1 - u^2)\,du = \sin x - \frac{\sin^3 x}{3} + C$.

Squares use the double angle. Odd powers use a Pythagorean identity and a substitution.

**Volumes.** Turning $y = \sin x$ for $0 \leq x \leq \pi$ about the $x$-axis gives $V = \pi\displaystyle\int_0^{\pi}\sin^2 x\,dx = \frac{\pi^2}{2}$. The region between $y = \cos x$ and $y = \sin x$ for $0 \leq x \leq \frac{\pi}{4}$ gives $V = \pi\displaystyle\int_0^{\pi/4}(\cos^2 x - \sin^2 x)\,dx = \pi\int_0^{\pi/4}\cos 2x\,dx = \frac{\pi}{2}$.

## Quick summary

- $\sin^2 x = \dfrac{1}{2}(1 - \cos 2x)$ and $\cos^2 x = \dfrac{1}{2}(1 + \cos 2x)$.
- $\displaystyle\int \sin^2 x\,dx = \frac{x}{2} - \frac{\sin 2x}{4} + C$ and $\displaystyle\int \cos^2 x\,dx = \frac{x}{2} + \frac{\sin 2x}{4} + C$.
- $\sin x\cos x = \dfrac{1}{2}\sin 2x$ and $\tan^2 x = \sec^2 x - 1$.
- For $\sin^2 kx$, the new angle is $2kx$.
- Odd powers: keep one factor and substitute.

## Quiz

Try these questions from the chapter quiz.

1. Which identity rewrites $\cos^2 x$?
   - A. $\dfrac{1}{2}(1 - \cos 2x)$
   - B. $\dfrac{1}{2}(1 + \cos 2x)$
   - C. $1 - \sin 2x$
   - D. $2\cos 2x - 1$
   - Answer: B. $\dfrac{1}{2}(1 + \cos 2x)$. From $\cos 2x = 2\cos^2 x - 1$.

2. Which identity rewrites $\sin^2 x$?
   - A. $\dfrac{1}{2}(1 - \cos 2x)$
   - B. $\dfrac{1}{2}(1 + \cos 2x)$
   - C. $\dfrac{1}{2}\sin 2x$
   - D. $1 + \cos^2 x$
   - Answer: A. $\dfrac{1}{2}(1 - \cos 2x)$. From $\cos 2x = 1 - 2\sin^2 x$.

3. Find $\displaystyle\int \cos 2x\,dx$.
   - A. $2\sin 2x + C$
   - B. $\dfrac{1}{2}\sin 2x + C$
   - C. $-\dfrac{1}{2}\sin 2x + C$
   - D. $\sin 2x + C$
   - Answer: B. $\dfrac{1}{2}\sin 2x + C$. Divide by the 2 inside.

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