8.2.7 Integrating with Trigonometric Identities

Step-by-step method:

infoNote

To integrate using trigonometric identities:

Simplify the expression with identities like half-angle, double-angle, or product-to-sum.
Apply standard integrals (e.g., $\int \sin(x) \, dx = -\cos(x) + C$ ).
Simplify the result and add the constant $C$ . This process reduces complex trigonometric integrals to manageable forms.

Trigonometric Integration

Integrating trig functions requires in-depth knowledge and quick recall of trigonometric identities.

Key Trigonometric Identities:

infoNote

\sin^2 \theta + \cos^2 \theta \equiv 1

\tan^2 \theta + 1 \equiv \sec^2 \theta

1 + \cot^2 \theta \equiv \csc^2 \theta

\cos 2\theta \equiv \cos^2 \theta - \sin^2 \theta = 2 \cos^2 \theta - 1 = 1 - 2 \sin^2 \theta

\sin 2\theta \equiv 2 \sin \theta \cos \theta

\tan \theta \equiv \frac{\sin \theta}{\cos \theta}

Examples of Trigonometric Integration:

infoNote

Example: $\int \tan^2 \theta \, d\theta$

(\tan^2 \theta +1 \equiv \sec^2 \theta)

=\int \sec^2\theta-1 \,d\theta \Rightarrow \tan \theta - \theta+ c

Example: $\int \tan \theta \, d\theta = \int \frac{\sin \theta}{\cos \theta} \, d\theta \quad (\tan \theta \equiv \frac {\sin \theta}{\cos \theta})$ Let $u = \cos \theta$

\Rightarrow \frac{du}{d\theta} = -\sin \theta \Rightarrow d\theta = \frac{-1}{\sin \theta} \, du

\Rightarrow I = \int \frac{\cancel\sin \theta}{u} \times \frac{-1}{\cancel\sin \theta} \, du = -\int \frac{1}{u} \, du = -\ln |u|

= -\ln |\cos \theta| + c \quad \text{(or } -1 \ln |\cos \theta | =\ln |(\cos \theta)^{-1} | = \ln |\sec \theta| \text{)}

Example: $\int \sin \theta \cos \theta \, d\theta = \int \frac{1}{2} \sin 2\theta \, d\theta$

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

Example: $\int \sin \theta \cos \theta \, d\theta$

\Rightarrow \text{Let } u = \sin \theta \Rightarrow \frac{du}{d\theta} = \cos \theta \Rightarrow d\theta = \frac{1}{\cos \theta} \, du

=\int u \,\cancel\cos \theta \times \frac{1}{\cancel\cos \theta} \, du

=\int u \, du = \frac {1}{2}u^2 + c = \frac {1}{2}\sin^2 \theta +c

Verification of Identity:

\frac{1}{2} \sin^2 \theta + \frac{c_1}{4} \cos 2\theta + c_2 = \frac{1}{4} \cos 2\theta + c_2

\Rightarrow \frac{1}{4} (1 - 2\sin^2 \theta) + c_2 = \frac{1}{4} + \frac{1}{2} \sin^2 \theta + c_2

= \frac{1}{2} \sin^2 \theta + \frac{1}{4}+c_2 \quad \checkmark

infoNote

Example

\int \tan^3 x \, dx = \int \tan^2 x \tan^2 x \, dx = \int \tan^2 x (\sec^2 x - 1) \, dx

= \int \tan^2 x \sec^2 x \quad(\text{\textcircled A}) - \int \tan^2 x \, dx \quad(\text{\textcircled B})

$\text{\textcircled A}$ : Let $u = \tan x$ $\Rightarrow \frac{du}{dx} = \sec^2 x \Rightarrow dx = \frac{1}{\sec^2 x} \, du$

\text{\textcircled A}\Rightarrow \int u\, \cancel\sec^2 x \times \frac {1}{\cancel\sec^2 x}\, du = \frac {1}{3}u^3+c = \frac {1}{3}\tan^3x+c

$\text{\textcircled B}$ :

\int \tan^2 x \, dx = \int \sec^2x-1 \, dx= \tan x - x +c_2

\therefore \quad I = \frac{1}{3}\tan^3 x - \tan x +x+d \leftarrow \text{integration constant}

Integrating $\sin^2 x \ and\ \cos^2 x$

infoNote

Expand $\cos 2x$ and write only in terms of what we are asked to integrate.
Make our integral the subject.
Replace our integral with the new integrable expression.

infoNote

Example:

\int \sin^2 x \, dx

\cos 2x = 1 - 2 \sin^2 x \\ 2\sin^2x = 1-\cos 2x\\ \sin^2 x = \frac{1}{2}-\frac {1}{2}\cos2x

I = \int \frac{1}{2} - \frac{1}{2} \cos (2x) \, dx

= \frac{1}{2}x - \frac{1}{4} \sin(2x) + c

infoNote

Example:

\int \cos^2 x \, dx

\cos 2x = 2 \cos^2 x - 1 \\ \Rightarrow 2\cos^2x=\cos2x +1\\ \Rightarrow \cos^2x=\frac {1}{2}\cos2x+\frac {1}{2}

= \int \frac {1}{2}\cos2x+\frac {1}{2}\, dx

= \frac{1}{2}x + \frac{1}{4} \sin(2x) + c

infoNote

Example:

\int \cos^2 (10x) \, dx

$\cos 2\theta = 2\cos^2 \theta -1$

Multiply by 2 and Divided by 2

\cos 20x = 2 \cos^2(10x) - 1\\ \Rightarrow 2\cos^2(10x) = \cos(20x) +1

\Rightarrow \cos^2(10x) = \frac{1 }{2}\cos (20x)+\frac {1}{2}

=\int \frac{1 }{2}\cos (20x)+\frac {1}{2} \, dx \\ = \frac {1}{40}\sin (20x)+\frac {1}{2}x+c

\int (x - \sin 2x)^2 \, dx = \int (x -\sin 2x)(x- \sin 2x) \, dx

= \int x^2-x\sin2x-x\sin2x+\sin^22x\, dx

=\int x^2 - 2x\sin2x + \sin^22x \, dx

\boxed {\text{\textcircled A} = x^2 \bigg | \text{\textcircled B} = 2x\sin2x \bigg | \text{\textcircled C}= \sin^22x \, dx}

$\text{\textcircled A}$ $\int x^2 \, dx$

= \frac{1}{3}x^3 + c

$\text{\textcircled B}$ $\int 2x \sin 2x \, dx$ By parts: $u = 2x$

\Rightarrow \frac{du}{dx} = 2

v = -\frac{1}{2} \cos 2x \quad \Rightarrow \quad \frac{dv}{dx} = \sin 2x

\Rightarrow I = -x \cos 2x - \int -\cos 2x \, dx

= -x \cos 2x + \int \cos 2x \, dx

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

$\text{\textcircled C}$ $\int \sin^2 2x \, dx$

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

= \int \frac {1}{2}-\frac {1}{2}\cos(4x)\, dx\\

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

\therefore \quad I = \frac{1}{3}x^3 - x \cos 2x - \frac{1}{2} \sin 2x + \frac{1}{2}x - \frac{1}{8}\sin 4x + c

Integrating with Trigonometric Identities Simplified Revision Notes for A-Level AQA Maths Pure

8.2.7 Integrating with Trigonometric Identities

Step-by-step method:

Trigonometric Integration

Examples of Trigonometric Integration:

Example

Integrating $\sin^2 x \ and\ \cos^2 x$

500K+ Students Use These Powerful Tools to Master Integrating with Trigonometric Identities For their A-Level Exams.

Other Revision Notes related to Integrating with Trigonometric Identities you should explore

Integrating with Trigonometric Identities Simplified Revision Notes for A-Level AQA Maths Pure

8.2.7 Integrating with Trigonometric Identities

Step-by-step method:

Trigonometric Integration

Examples of Trigonometric Integration:

Example

Integrating sin⁡2x and cos⁡2x\sin^2 x \ and\ \cos^2 xsin2x and cos2x

500K+ Students Use These Powerful Tools to Master Integrating with Trigonometric Identities For their A-Level Exams.

Other Revision Notes related to Integrating with Trigonometric Identities you should explore

Integrating $\sin^2 x \ and\ \cos^2 x$