hw7sol - x ) | x = π/ 4 x =0 = ± 1 + √ 2 ²-(0 + 1) =...

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Fall 2010, MAT21B, Solution to Paper Homework 7 1. Evaluate the integral Z e 2 x cos(3 x ) dx Solution: Let u = e 2 x ,dv = cos (3 x ) dx du = 2 e 2 x dx,v = 1 3 sin (3 x ) By the method of integration by parts, we have Z e 2 x cos(3 x ) dx = 1 3 e 2 x sin (3 x ) - 2 3 Z sin (3 x ) e 2 x dx Now, we use the method of integration by parts again, let u = e 2 x ,dv = sin (3 x ) dx du = 2 e 2 x dx,v = - 1 3 cos (3 x ) Then, Z sin (3 x ) e 2 x dx = - 1 3 e 2 x cos (3 x ) + 2 3 Z cos (3 x ) e 2 x dx Therefore, Z e 2 x cos(3 x ) dx = 1 3 e 2 x sin (3 x ) - 2 3 Z sin (3 x ) e 2 x dx = 1 3 e 2 x sin (3 x ) - 2 3 ± - 1 3 e 2 x cos (3 x ) + 2 3 Z cos (3 x ) e 2 x dx ² = 1 3 e 2 x sin (3 x ) + 2 9 e 2 x cos (3 x ) - 4 9 Z cos (3 x ) e 2 x dx So, 13 9 Z e 2 x cos(3 x ) dx = 1 3 e 2 x sin (3 x ) + 2 9 e 2 x cos (3 x ) + C Z e 2 x cos(3 x ) dx = 3 13 e 2 x sin (3 x ) + 2 13 e 2 x cos (3 x ) + C 1
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2. Evaluate the integral Z π 4 0 1 + sin x cos 2 x dx Solution: Z π 4 0 1 + sin x cos 2 x dx = Z π 4 0 1 cos 2 x + sin x cos 2 x dx = Z π 4 0 sec 2 x + sec x tan xdx = (tan x + sec
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Unformatted text preview: x ) | x = π/ 4 x =0 = ± 1 + √ 2 ²-(0 + 1) = √ 2 . 3. Evaluate the integral Z x (ln x ) 2 dx Solution: Let u = (ln x ) 2 ,dv = xdx Then, du = 2 ln x x dx and v = x 2 2 . By using integration by parts, we have Z x (ln x ) 2 dx = x 2 2 (ln x ) 2-Z x 2 2 . 2 ln x x dx = x 2 2 (ln x ) 2-Z x ln xdx = x 2 2 (ln x ) 2-³ x 2 2 ln x-Z x 2 2 . 1 x dx ´ = x 2 2 (ln x ) 2-x 2 2 ln x + Z x 2 dx = x 2 2 (ln x ) 2-x 2 2 ln x + x 2 4 + C 2...
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This note was uploaded on 10/08/2010 for the course MATH 21B taught by Professor Vershynin during the Spring '08 term at UC Davis.

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hw7sol - x ) | x = π/ 4 x =0 = ± 1 + √ 2 ²-(0 + 1) =...

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