Homework 7

# Homework 7 - Rehman(aar638 – HW07 – sachse –(56620 1...

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Unformatted text preview: Rehman (aar638) – HW07 – sachse – (56620) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Evaluate the definite integral I = integraldisplay π/ 2 (5 sin θ + 2 sin 3 θ ) dθ . 1. I = 3 2. I = 19 3 correct 3. I = 17 3 4. I = 11 3 5. I = 7 6. I = 13 3 Explanation: Since sin 2 θ = 1- cos 2 θ we see that 5 sin θ + 2 sin 3 θ = sin θ (5 + 2 sin 2 θ ) = sin θ [5 + 2 (1- cos 2 θ )] = sin θ (7- 2 cos 2 θ ) . Thus I = integraldisplay π/ 2 sin θ (7- 2 cos 2 θ ) dθ As the integral is now of the form sin θ f (cos θ ) , f ( x ) = 7- 2 x 2 , the subsitution x = cos θ is suggested. For then dx =- sin θ dθ , while θ = 0 = ⇒ x = 1 , θ = π 2 = ⇒ x = 0 . In this case I =- integraldisplay 1 (7- 2 x 2 ) dx = integraldisplay 1 (7- 2 x 2 ) dx . Consequently, I = bracketleftBig 7 x- 2 3 x 3 bracketrightBig 1 = 19 3 . 002 10.0 points Evaluate the integral I = integraldisplay π/ 2 3 sin 3 x cos 2 x dx . 1. I = 8 5 2. I = 4 5 3. I = 2 5 correct 4. I = 6 5 5. I = 1 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 . For then du =- sin x dx , while x = 0 = ⇒ u = 1 x = π 2 = ⇒ u = 0 . Rehman (aar638) – HW07 – sachse – (56620) 2 In this case I =- integraldisplay 1 3( u 2- u 4 ) du . Consequently, I = bracketleftBig- u 3 + 3 5 u 5 bracketrightBig 1 = 2 5 . keywords: Stewart5e, indefinite integral, powers of sin, powers of cos, trig substitu- tion, 003 10.0 points Evaluate the definite integral I = integraldisplay π/ 4 5 cos x + 4 sin x cos 3 x dx . 1. I = 11 2 2. I = 7 correct 3. I = 6 4. I = 5 5. I = 13 2 Explanation: After division 5 cos x + 4 sin x cos 3 x = 5 sec 2 x + 4 tan x sec 2 x = (5 + 4 tan x ) sec 2 x . Thus I = integraldisplay π/ 4 (5 + 4 tan x ) sec 2 x dx . Let u = tan x ; then du = sec 2 x dx so I = integraldisplay 1 (5 + 4 u ) du = bracketleftbig 5 u + 2 u 2 bracketrightbig 1 ....
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Homework 7 - Rehman(aar638 – HW07 – sachse –(56620 1...

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