242hwsolns0304 - Brief solutions for(parts of Homeworks 3...

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Unformatted text preview: Brief solutions for (parts of) Homeworks 3 and 4 Homework 4 1. Z π/ 6 cos 3 x dx = Z π/ 6 (1- sin 2 x ) cos x dx = Z sin π/ 6 sin0 (1- u 2 ) du = h u- u 3 3 i 1 / 2 = 1 2- 1 24 = 11 24 . 2. I = Z sec 3 x dx = Z sec x sec 2 x dx . Integrate by parts with u = sec x , dv = sec 2 x dx , so du = sec x tan x dx and v = tan x . Then I = sec x tan x- Z sec x tan 2 x dx I = sec x tan x- Z sec x (sec 2 x- 1) dx I = sec x tan x- Z sec 3 x dx + Z sec x dx I = sec x tan x- I + ln | sec x + tan x | 2 I = sec x tan x + ln | sec x + tan x | I = 1 2 sec x tan x + ln | sec x + tan x | + C 3. Z x 3 √ x 2 + 100 dx = Z 1000 tan 3 θ √ 100 tan 2 θ + 100 10 sec 2 θ dθ = Z 1000 tan 3 θ 10 sec θ 10 sec 2 θ dθ = 1000 Z tan 3 θ sec θ dθ = 1000 Z tan 2 θ sec θ tan θ dθ = 1000 Z (sec 2 θ- 1) sec θ tan θ dθ = 1000 Z ( u 2- 1) du = 1000 u 3 3- u = 1000 sec 3 θ 3- sec θ . 1 Since x 10 = tan θ , it follows that √ x 2 + 100 10 = sec θ . The final answer is then 1000 1 3 · ( x 2 + 100) 3 /...
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This note was uploaded on 12/07/2011 for the course MATH 242 taught by Professor Wang during the Spring '08 term at University of Delaware.

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242hwsolns0304 - Brief solutions for(parts of Homeworks 3...

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