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

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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
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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.

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

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