## Calculus Of One Real Variable – By Pheng Kim Ving Chapter 10: Techniques Of Integration – Section 10.3: Integration Of Trigonometric Fumctions

10.3
Integration Of Trigonometric Functions

Go To Problems & Solutions

 1. Basic Trigonometric Integrals

Recall from Section 10.1 Part 4 that:

For sec x:

We group the basic trigonometric integrals together here in the following box.

Evaluate:

## SolutionEOS

 2. Trigonometric Substitution

# Example 2.1

Evaluate:

## Solution

Let u = sin 3x. Then du = 3 cos 3x dx, so that cos 3x dx = (1/3) du. Thus:

## EOS

The substitution u = sin 3x involves a trigonometric function, and as a consequence is called a trigonometric
substitution
.

Find:

## Solution

Let u = x2. Then du = 2x dx, so that x dx = (1/2) du. Thus:

EOS

The substitution u = x2 doesn't involve any trigonometric function. There's no trigonometric substitution. Integrals
involving trigonometric functions aren't always handled by using a trigonometric substitution.

Note that sin x2 = sin (x2), the sine of x2, not (sin x)2, denoted sin2 x, the square of sin x.

 Problems And Solutions

1. Evaluate:

Solution

Let u = 1 + sin x. So du = cos x dx. Thus:

2. Calculate:

Solution

Let u = ln t. So du = (1/t) dt. Thus:

3. Compute:

Solution

Let v = 2 + sin 3u. Then dv = 3 cos 3u du, so that cos 3u du = (1/3) dv. Thus:

4. Find:

Solution

5. a. Establish the following identities:

Solution

a. We have cos (xy) = cos x cos y + sin x sin y and cos (x + y) = cos x cos ysin x sin y. So
cos (xy) – cos (x + y) = 2 sin x sin y, which yields: