FA06Ma.Hw8solutions

FA06Ma.Hw8solutions - 5. 5.10.14 (p221) Use integration by...

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Math 1a, Section 1 Solutions to Homework 8 1. Q1 - See solutions in the book. 2. Q2 - See solutions in the book. 3. 5.8.3, 5.8.14, 5.8.18 (p216) (a) Let y = 1 + x . Then x 2 = ( y - 1) 2 and dx = dy . Hence, Z x 2 1 + xdx = Z ( y - 1) 2 ydy = y 7 / 2 / (7 / 2) - 2 y 5 / 2 / (5 / 2) + y 3 / 2 / (3 / 2) = (1 + x ) 7 / 2 / (7 / 2) - 2(1 + x ) 5 / 2 / (5 / 2) + (1 + x ) 3 / 2 / (3 / 2) (1) (b) Let y = 1 - x 6 . Then dy/dx = - 6 x 5 . Hence Z x 5 1 - x 6 dx = Z dy - 6 y = - 1 / 3(1 - x 6 ) 1 / 2 (2) (c) Let y = sin x - cos x . Then dy/dx = cos x + sin x . Hence Z sin x + cos x (sin x - cos x ) 1 / 3 dx = Z y - 1 / 3 dy = 3(sin x - cos x ) 2 / 3 2 (3) 4. 5.10.7 (p220) Z sin 2 xdx = Z sin x d ( - cos x ) = - cos x sin x - Z ( - cos x ) d sin x = - cos x sin x + Z cos 2 xdx (4) Apply cos 2 x = 1 - sin 2 x to the right hand side of (4). Then we have 2 Z sin 2 xdx = - sin x cos x + x (5) Furthermore, sin(2 x ) = 2 sin x cos x and the result follows.
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Unformatted text preview: 5. 5.10.14 (p221) Use integration by parts to obtain Z √ 1-x 2 dx = x √ 1-x 2-Z xd ( √ 1-x 2 ) = x √ 1-x 2-Z x ±-2 x 2 √ 1-x 2 ² dx = x √ 1-x 2 + Z x 2 dx √ 1-x 2 (6) Since x 2 = x 2-1 + 1, Z x 2 dx √ 1-x 2 = Z x 2-1 √ 1-x 2 + 1 √ 1-x 2 dx =-Z √ 1-x 2 dx + Z 1 √ 1-x 2 dx (7) Substituting (7) into (6), we arrive at the following and the result follows. 2 Z √ 1-x 2 dx = x √ 1-x 2 + Z 1 √ 1-x 2 dx (8)...
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This note was uploaded on 12/15/2009 for the course MA 1a taught by Professor Borodin,a during the Spring '08 term at Caltech.

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FA06Ma.Hw8solutions - 5. 5.10.14 (p221) Use integration by...

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