Trigonometric Functions

Step-by-step method:

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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:

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\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:

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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

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

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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.

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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

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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

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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

Trigonometric Functions Simplified Revision Notes for Leaving Cert Mathematics

Trigonometric Functions

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 Trigonometric Functions For their Leaving Cert Exams.

Other Revision Notes related to Trigonometric Functions you should explore

Trigonometric Functions Simplified Revision Notes for Leaving Cert Mathematics

Trigonometric Functions

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 Trigonometric Functions For their Leaving Cert Exams.

Other Revision Notes related to Trigonometric Functions you should explore

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