8.4.15-eg-1

# 8.4.15-eg-1 - Example Evaluate: 1 cos x dx. 0 Two solutions...

This preview shows pages 1–2. Sign up to view the full content.

Example Evaluate: Z π 0 1 - cos xdx . Two solutions are shown below: 1. First solution: Using the double angle formula, sin 2 θ = 1 - cos 2 θ 2 , or in this case, 2 sin 2 ± x 2 ² = 1 - cos x. Thus, Z π 0 1 - cos xdx = Z π 0 s 2 sin 2 ± x 2 ² dx = 2 Z π 0 ³ ³ ³ ³ sin ± x 2 ²³ ³ ³ ³ dx, using the fact that q y 2 = | y | . Then since sin ´ x 2 µ 0 for x [0 ], Z π 0 1 - cos xdx = 2 Z π 0 sin ± x 2 ² dx = - 2 2 cos ± x 2 ²³ ³ ³ ³ π 0 = - 2 2 (0 - 1) = 2 2 . 2. Second solution: Z π 0 1 - cos xdx = Z π 0 1 - cos x 1 · 1 + cos x 1 + cos x dx = Z π 0 1 - cos 2 x 1 + cos x dx = Z π 0

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 04/02/2012 for the course MTH 132 taught by Professor Kihyunhyun during the Fall '10 term at Michigan State University.

### Page1 / 2

8.4.15-eg-1 - Example Evaluate: 1 cos x dx. 0 Two solutions...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online