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solution_pdf7 - Horstman(mdh995 HW07 Radin(58505 This...

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Horstman (mdh995) – HW07 – Radin – (58505) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points Evaluate the defnite integral I = i π/ 2 0 (5 sin θ - 3 sin 3 θ ) dθ . 1. I = 8 2. I = 2 3. I = 4 4. I = 7 5. I = 3 correct 6. I = 6 Explanation: Since sin 2 θ = 1 - cos 2 θ we see that 5 sin θ - 3 sin 3 θ = sin θ (5 - 3 sin 2 θ ) = sin θ [5 - 3 (1 - cos 2 θ )] = sin θ (2 + 3 cos 2 θ ) . Thus I = i π/ 2 0 sin θ (2 + 3 cos 2 θ ) As the integral is now oF the Form sin θ f (cos θ ) , f ( x ) = 2 + 3 x 2 , the subsitution x = cos θ is suggested. ±or then dx = - sin θ dθ , while θ = 0 = x = 1 , θ = π 2 = x = 0 . In this case I = - i 0 1 (2 + 3 x 2 ) dx = i 1 0 (2 + 3 x 2 ) dx . Consequently, I = b 2 x + x 3 B 1 0 = 3 . 002 10.0 points Evaluate the integral I = i π/ 2 0 3 sin 3 x cos 2 x dx . 1. I = 2 5 correct 2. I = 8 5 3. I = 1 5 4. I = 4 5 5. I = 6 5 Explanation: Since sin 3 x cos 2 x = sin x (sin 2 x cos 2 x ) = sin x (1 - cos 2 x )cos 2 x = sin x (cos 2 x - cos 4 x ) , the integrand is oF the Form sin xf (cos x ), sug- gesting use oF the substitution u = cos x . ±or then du = - sin x dx , while x = 0 = u = 1 x = π 2 = u = 0 .
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Horstman (mdh995) – HW07 – Radin – (58505) 2 In this case I = - i 0 1 3( u 2 - u 4 ) du . Consequently, I = b - u 3 + 3 5 u 5 B 0 1 = 2 5 . keywords: Stewart5e, indefnite integral, powers oF sin, powers oF cos, trig substitu- tion, 003 10.0 points Evaluate the indefnite integral I = i 1 - sin x cos x dx . 1. I = - ln(1 - cos x ) + C 2. I = ln(1 + cos x ) + C 3. I = ln(1 - sin x ) + C 4. I = ln(1 + sin x ) + C correct 5. I = - ln(1 + cos x ) + C Explanation: i 1 - sin x cos x dx = i (sec x - tan x ) dx = ln | sec x + tan x | - ln | sec x | + C = ln | (sec x + tan x ) cos x | + C = ln(1 + sin x ) + C 004 10.0 points Evaluate the defnite integral I = i π/ 3 0 sec x tan x 4 + sec x dx . 1. I = ln 6 5 correct 2. I = ln 5 4 3. I = - ln 5 4 4. I = ln 3 4 5. I = - ln 6 5 6. I = - ln 3 4 Explanation: Since d dx (sec x ) = sec x tan x , use oF the substitution u = 4 + sec x is suggested. ±or then du = sec x tan xdx , while x = 0 = u = 5 , x = π 3 = u = 6 . Thus I = i 6 5 1 u du = b ln u B 6 5 . Consequently, I = ln 6 5 . 005 10.0 points Evaluate the defnite integral I = i π/ 4 0 4 tan 4 x dx . 1. I = 4 π - 4 3
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Horstman (mdh995) – HW07 – Radin – (58505) 3 2. I = 4 π + 4 3 3. I = 2 π - 8 3 4. I = 2 π + 4 3 5. I = π + 8 3 6. I = π - 8 3 correct Explanation: Since tan 2 x = sec 2 x - 1 , it follows that tan 4 x = tan 2 x (sec 2 x - 1) = tan 2 x sec 2 x - tan 2 x . Thus, using the same trig identity as before, we see that tan 4 x = (tan 2 x - 1) sec 2 x + 1 , in which case I = 4 i π/ 4 0 b (tan 2 x - 1) sec 2 x + 1 B dx . The whole point of this use of trig identities is that d dx tan x = sec 2 x , so set u = tan x . Then du = sec 2 x dx , while x = 0 = u = 0 , x = π 4 = u = 1 .
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