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8.3 Trig Integrals (part 1)

NOT on exam 1

integrals involving cosine, sine, tangent, secant

\[ \int \frac{\sin x}{\cos x} dx \]

Let \( u = \cos x \), then \( du = -\sin x dx \)

\[ = \int \frac{-1}{u} du = -\ln |u| + C = -\ln |\cos x| + C \]

basic idea:

remember \( \cos x \) and \( \sin x \) are related by derivative \( \rightarrow \) substitution possible if they both show up.

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example

\[ \int \frac{1}{1 - \sin x} dx \]

can't integrate directly, bring \( \cos x \) in somehow?

\[ = \int \frac{1}{1 - \sin x} \cdot \frac{1 + \sin x}{1 + \sin x} dx \]\[ = \int \frac{1 + \sin x}{1 - \sin^2 x} dx \]

\( \cos^2 x + \sin^2 x = 1 \)\\ \( \cos^2 x = 1 - \sin^2 x \)

\[ = \int \frac{1 + \sin x}{\cos^2 x} dx = \int \frac{1}{\cos^2 x} dx + \int \frac{\sin x}{\cos^2 x} dx \]

Term 1:

\[ \sec^2 x \]

Term 2 Substitution:

\( u = \cos x \), \( du = -\sin x dx \)

\[ = \int \sec^2 x dx - \int \frac{1}{u^2} du \]\[ = \tan x + \frac{1}{u} + C = \tan x + \frac{1}{\cos x} + C \]

\[ \tan x + \frac{1}{\cos x} + C \]

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Example: Definite Integral with Trigonometric Substitution

Evaluate the following integral:

\[ \int_{-\pi/2}^{0} \sqrt{1 + \cos(2x)} \, dx \]

Somehow bring sine into this.

\[ = \int_{-\pi/2}^{0} \sqrt{\frac{1 + \cos(2x)}{1} \cdot \frac{1 - \cos(2x)}{1 - \cos(2x)}} \, dx \]

\[ = \int_{-\pi/2}^{0} \sqrt{\frac{1 - \cos^2(2x)}{1 - \cos(2x)}} \, dx = \int_{-\pi/2}^{0} \sqrt{\frac{\sin^2(2x)}{1 - \cos(2x)}} \, dx \]

\[ = \int_{-\pi/2}^{0} \frac{|\sin(2x)|}{\sqrt{1 - \cos(2x)}} \, dx \]

\[ |x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases} \]

Interval Analysis

Interval: \( -\frac{\pi}{2} \le x \le 0 \)

\( \sin(2x) < 0 \) on this interval.

So, \( |\sin(2x)| = -\sin(2x) \)

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Substituting the absolute value result back into the integral:

\[ \int_{-\pi/2}^{0} \frac{-\sin(2x)}{\sqrt{1 - \cos(2x)}} \, dx \]

Substitution Method

Let \( u = 1 - \cos(2x) \)

Then \( du = 2 \sin(2x) \, dx \)

⋮

\[ \dots = \sqrt{2} \]

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Example: Trigonometric Integrals

\[ \int \sin^2 x \cos^5 x \, dx \]

Basic idea: let \( u = \cos x \) or \( u = \sin x \) and see which works.

Strategy for \( \int \sin^m x \cos^n x \, dx \)

Case 1: if \( m \) or \( n \) is positive and odd

Then split one power of the part w/ odd power and save it, then use \( \sin^2 x + \cos^2 x = 1 \) to turn everything into the other one.

Case 2: if \( m \) and \( n \) are both positive and even

Then use:

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

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

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Back to \( \int \sin^2 x \cos^5 x \, dx \)

Here \( \cos x \) has positive odd power. Save a factor of \( \cos x \).

\[ \int \sin^2 x \cos^4 x \cos x \, dx \]

Now turn everything into \( \sin x \):

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

\[ = \int \sin^2 x (1 - \sin^2 x)^2 \cos x \, dx \]

\( u = \sin x \)

\( du = \cos x \, dx \)

the factor of \( \cos x \) saved

\[ = \int u^2 (1 - u^2)^2 \, du \]

\[ = \dots = \frac{1}{3} \sin^3 x - \frac{2}{5} \sin^5 x + \frac{1}{7} \sin^7 x + C \]

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Example: Integral of Even Powers of Sine and Cosine

\[ \int \sin^2 x \cos^2 x \, dx \]

Both even power: use

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

\[ = \int \frac{1 - \cos(2x)}{2} \cdot \frac{1 + \cos(2x)}{2} \, dx \]\[ = \int \frac{1 - \cos^2(2x)}{4} \, dx = \frac{1}{4} \int (1 - \cos^2(2x)) \, dx \]\[ = \frac{1}{4} \int \sin^2(2x) \, dx \]

Use:

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

\[ = \frac{1}{4} \int \frac{1 - \cos(4x)}{2} \, dx \]\[ = \frac{1}{8} \int (1 - \cos(4x)) \, dx \]

\[ = \frac{1}{8} \left( x - \frac{1}{4} \sin(4x) \right) + C \]

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Brief Look at \(\tan x\) and \(\sec x\)

Basic idea: substitution knowing

\[ \frac{d}{dx} \tan x = \sec^2 x \]\[ \frac{d}{dx} \sec x = \sec x \tan x \]\[ \text{and } \tan^2 x + 1 = \sec^2 x \]

Example

\[ \int \tan^3 x \, dx \]

We see \(\tan x\), we would like to have \(\sec^2 x\) somewhere.

\[ \int \tan x \cdot \tan^2 x \, dx = \int \tan x (\sec^2 x - 1) \, dx \]

Note: \(\tan^2 x = \sec^2 x - 1\)

\[ = \int \tan x \cdot \sec^2 x \, dx - \int \tan x \, dx = \dots \]

First part:

\(u = \tan x\)

\(du = \sec^2 x \, dx\)

Second part:

\(\tan x = \frac{\sin x}{\cos x}\)

\(u = \cos x\), \(du = -\sin x \, dx\)

(see first example)

\[ = \frac{1}{2} \tan^2 x + \ln |\cos x| + C \]

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Trigonometric Derivatives and Identities

\[ \frac{d}{dx} \cot x = -\csc^2 x \]

\[ \frac{d}{dx} \csc x = -\csc x \cot x \]

\[ 1 + \cot^2 x = \csc^2 x \]

Example

Evaluate the following integral:

\[ \begin{aligned} &\int \cot^2 x \, dx \\ &= \int (\csc^2 x - 1) \, dx \\ &= \int \csc^2 x \, dx - \int dx \end{aligned} \]

= -\cot x - x + C

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SI Exam Review

Mon. 9/18 5:30 - 7:30 pm

MATH 175