242hwsolns0304

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

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Brief solutions for (parts of) Homeworks 3 and 4 Homework 4 1. Z π/ 6 0 cos 3 x dx = Z π/ 6 0 (1 - sin 2 x ) cos x dx = Z sin π/ 6 sin 0 (1 - u 2 ) du = h u - u 3 3 i 1 / 2 0 = 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

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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 / 2 1000 - x 2 + 100 10 + C which can probably be written in other ways as well. 4. We look for A and B that satisfy x - 9 x 2 + 3 x - 10 = x - 9 ( x - 2)( x + 5) = A x - 2 + B x + 5 which implies x - 9 = A ( x +5)+ B ( x - 2). Either expanding and comparing,
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