Using Trigonometric Identities for Integration

Transforming expressions we can't integrate into ones we can

SPEC 8.2 • INTEGRATION

The Problem

We know how to integrate basic trig functions:

What We Can Integrate

\int sin x d x = - cos x + c

\int cos x d x = sin x + c

\int sec^{2} x d x = tan x + c

\int csc^{2} x d x = - cot x + c

The Challenge

But what about

\int sin^{2} x d x

,

\int cos^{2} x d x

,

\int tan^{2} x d x

, or

\int cot^{2} x d x

?

There's NO direct antiderivative! We need a different approach.

Solution

Use trigonometric identities to rewrite these expressions into forms we CAN integrate!

Essential Identities

These five identities are essential for integration:

FunctionIdentityRearranged for Integration

cos^{2} x

cos 2 x = 2 cos^{2} x - 1

cos^{2} x = \frac{1}{2} (1 + cos 2 x)

sin^{2} x

cos 2 x = 1 - 2 sin^{2} x

sin^{2} x = \frac{1}{2} (1 - cos 2 x)

tan^{2} x

1 + tan^{2} x = sec^{2} x

tan^{2} x = sec^{2} x - 1

cot^{2} x

1 + cot^{2} x = csc^{2} x

cot^{2} x = csc^{2} x - 1

sin x cos x

sin 2 x = 2 sin x cos x

sin x cos x = \frac{1}{2} sin 2 x

Memory Tip

cos²x: "cos goes with PLUS" → ½(1 + cos 2x)

sin²x: "sin goes with MINUS" → ½(1 − cos 2x)

The Method

3-Step Process

Step 1: Identify which identity to use

Step 2: Substitute the identity and simplify

Step 3: Integrate term by term

For sin² and cos²

Replace with the double angle form, then integrate:

\int cos^{2} x d x = \int \frac{1}{2} (1 + cos 2 x) d x = \frac{x}{2} + \frac{s i n 2 x}{4} + c

\int sin^{2} x d x = \int \frac{1}{2} (1 - cos 2 x) d x = \frac{x}{2} - \frac{s i n 2 x}{4} + c

The same method applies for sin²(kx) and cos²(kx) — just adjust the coefficient when integrating.

For tan² and cot²

Replace with the Pythagorean identity form, then integrate:

\int tan^{2} x d x = \int (sec^{2} x - 1) d x = tan x - x + c

\int cot^{2} x d x = \int (csc^{2} x - 1) d x = - cot x - x + c

Worked Example01

Find $\int sin^{2} 3 x d x$

1

Use identity:

sin^{2} (3 x) = \frac{1}{2} (1 - cos 6 x)

2

\int sin^{2} 3 x d x = \int \frac{1}{2} (1 - cos 6 x) d x

3

= \frac{1}{2} (x - \frac{s i n 6 x}{6}) + c

4

= \frac{x}{2} - \frac{s i n 6 x}{12} + c

Answer

\frac{x}{2} - \frac{s i n 6 x}{12} + c

Worked Example02

Find $\int tan^{2} x d x$

1

Use identity:

tan^{2} x = sec^{2} x - 1

2

\int tan^{2} x d x = \int (sec^{2} x - 1) d x

3

= tan x - x + c

Answer

tan x - x + c

Worked Example03

Find $\int sin x cos x d x$

1

Use identity:

sin x cos x = \frac{1}{2} sin 2 x

2

\int sin x cos x d x = \int \frac{1}{2} sin 2 x d x

3

= \frac{1}{2} \times (- \frac{1}{2} cos 2 x) + c

4

= - \frac{1}{4} cos 2 x + c

Answer

- \frac{1}{4} cos 2 x + c

Worked Example04

Evaluate $\int_{0}^{\frac{π}{2}} sin^{2} x d x$

1

Use identity:

sin^{2} x = \frac{1}{2} (1 - cos 2 x)

2

\int_{0}^{\frac{π}{2}} sin^{2} x d x = \int_{0}^{\frac{π}{2}} \frac{1}{2} (1 - cos 2 x) d x

3

= \frac{1}{2} [x - \frac{s i n 2 x}{2}]_{0}^{\frac{π}{2}}

4

= \frac{1}{2} [(\frac{π}{2} - \frac{s i n π}{2}) - (0 - \frac{s i n 0}{2})]

5

= \frac{1}{2} (\frac{π}{2} - 0 - 0 + 0) = \frac{π}{4}

Answer

\frac{π}{4}

Worked Example05

Find $\int sin^{2} x cos^{2} x d x$

1

Use:

sin x cos x = \frac{1}{2} sin 2 x

, so

sin^{2} x cos^{2} x = \frac{1}{4} sin^{2} 2 x

2

Now use:

sin^{2} 2 x = \frac{1}{2} (1 - cos 4 x)

3

sin^{2} x cos^{2} x = \frac{1}{4} \times \frac{1}{2} (1 - cos 4 x) = \frac{1}{8} (1 - cos 4 x)

4

\int sin^{2} x cos^{2} x d x = \frac{1}{8} (x - \frac{s i n 4 x}{4}) + c

5

= \frac{x}{8} - \frac{s i n 4 x}{32} + c

Answer

\frac{x}{8} - \frac{s i n 4 x}{32} + c

Practice

Question 1: Find

\int (1 + cos x)^{2} d x

Step 1: Expand:

(1 + cos x)^{2} = 1 + 2 cos x + cos^{2} x

Step 2: Use identity for cos²x:

= 1 + 2 cos x + \frac{1}{2} (1 + cos 2 x)

Step 3: Simplify:

= \frac{3}{2} + 2 cos x + \frac{1}{2} cos 2 x

Step 4: Integrate:

= \frac{3 x}{2} + 2 sin x + \frac{s i n 2 x}{4} + c

Answer

\frac{3 x}{2} + 2 sin x + \frac{s i n 2 x}{4} + c

Practice

Question 2: Find

\int sin^{4} x d x

Step 1: Write as

(sin^{2} x)^{2} = [\frac{1}{2} (1 - cos 2 x)]^{2}

Step 2: Expand:

= \frac{1}{4} (1 - 2 cos 2 x + cos^{2} 2 x)

Step 3: Use identity for cos²2x:

cos^{2} 2 x = \frac{1}{2} (1 + cos 4 x)

Step 4:

= \frac{1}{4} (1 - 2 cos 2 x + \frac{1}{2} + \frac{1}{2} cos 4 x) = \frac{1}{4} (\frac{3}{2} - 2 cos 2 x + \frac{1}{2} cos 4 x)

Step 5: Integrate:

= \frac{1}{4} (\frac{3 x}{2} - sin 2 x + \frac{s i n 4 x}{8}) + c

Answer

\frac{3 x}{8} - \frac{s i n 2 x}{4} + \frac{s i n 4 x}{32} + c

Practice

Question 3: Evaluate

\int_{0}^{\frac{π}{4}} tan^{2} x d x

Step 1: Use identity:

tan^{2} x = sec^{2} x - 1

Step 2:

\int_{0}^{\frac{π}{4}} tan^{2} x d x = \int_{0}^{\frac{π}{4}} (sec^{2} x - 1) d x

Step 3:

= [tan x - x]_{0}^{\frac{π}{4}}

Step 4:

= (tan \frac{π}{4} - \frac{π}{4}) - (tan 0 - 0)

Step 5:

= (1 - \frac{π}{4}) - 0 = 1 - \frac{π}{4}

Answer

1 - \frac{π}{4}

Summary

Key Integration Results

FunctionIntegral

cos^{2} x

\frac{x}{2} + \frac{s i n 2 x}{4} + c

sin^{2} x

\frac{x}{2} - \frac{s i n 2 x}{4} + c

tan^{2} x

tan x - x + c

cot^{2} x

- cot x - x + c

sin x cos x

- \frac{1}{4} cos 2 x + c

Key Exam Skills

Recognise when an identity is needed (can't integrate directly)

Choose the correct identity for the function

For higher powers like sin⁴x, apply identities multiple times

Always check by differentiating back if unsure