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Unformatted text preview: pokharel (yp624) – HW08 – Radin – (57410) 1 This printout should have 23 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Evaluate the integral I = integraldisplay π/ 2 sin 2 x cos 3 x dx . 1. I = 4 15 2. I = 1 15 3. I = 8 15 4. I = 2 15 correct 5. I = 2 5 Explanation: Since sin 2 x cos 3 x = (sin 2 x cos 2 x ) cos x = sin 2 x (1 sin 2 x )cos x = (sin 2 x sin 4 x )cos x , the integrand is of the form cos xf (sin x ), sug gesting use of the substitution u = sin x . For then du = cos x dx , while x = 0 = ⇒ u = 0 x = π 2 = ⇒ u = 1 . In this case I = integraldisplay 1 ( u 2 u 4 ) du . Consequently, I = bracketleftBig 1 3 u 3 1 5 u 5 bracketrightBig 1 = 2 15 . keywords: Stewart5e, indefinite integral, powers of sin, powers of cos, trig substitu tion, 002 10.0 points Evaluate the definite integral I = integraldisplay π/ 4 3 cos x 2 sin x cos 3 x dx . 1. I = 3 2 2. I = 2 correct 3. I = 5 2 4. I = 1 2 5. I = 1 Explanation: After division 3 cos x 2 sin x cos 3 x = 3 sec 2 x 2 tan x sec 2 x = (3 2 tan x ) sec 2 x . Thus I = integraldisplay π/ 4 (3 2 tan x ) sec 2 x dx . Let u = tan x ; then du = sec 2 x dx so I = integraldisplay 1 (3 2 u ) du = bracketleftbig 3 u u 2 bracketrightbig 1 . pokharel (yp624) – HW08 – Radin – (57410) 2 Consequently, I = 2 . 003 10.0 points Find the value of I = integraldisplay π 4 tan 4 x dx . 1. I = π 4 2 3 correct 2. I = π 3 3. I = π 6 8 √ 3 9 4. I = π 6 + 8 √ 3 9 5. I = π √ 3 3 6. I = π 4 + 2 3 Explanation: Since tan 2 x = sec 2 x 1 , we see that tan 4 x = tan 2 x ( sec 2 x 1 ) = tan 2 x sec 2 x tan 2 x . Thus by trig identities yet again, tan 4 x = ( tan 2 x 1 ) sec 2 x + 1 . In this case, I = integraldisplay π 4 bracketleftbig( tan 2 x 1 ) sec 2 x + 1 bracketrightbig dx = bracketleftbigg 1 3 tan 3 x tan x + x bracketrightbigg π 4 . On the other hand, tan π 4 = 1 . Consequently, I = π 4 2 3 . 004 10.0 points Evaluate the definite integral I = integraldisplay π x (5 cos 2 x sin 2 x ) dx 1. I = 5 2 π 2 2. I = 3 2 π 2 3. I = π 2 correct 4. I = π 2 + 1 5. I = 3 2 π 2 + 3 2 Explanation: Since cos 2 x = 1 2 (1 + cos 2 x ) and sin 2 x = 1 2 (1 cos 2 x ) , we see that I = 1 2 integraldisplay π x { 5 (1 + cos 2 x ) 1 + cos 2 x } dx = 2 integraldisplay π x dx + 3 integraldisplay π x cos 2 x dx = bracketleftBig x 2 bracketrightBig π + 3 integraldisplay π x cos 2 x dx = π 2 + 3 integraldisplay π x cos 2 x dx . pokharel (yp624) – HW08 – Radin – (57410) 3 But after integration by parts, integraldisplay π x cos 2 x dx = 1 2 bracketleftBig x sin 2 x bracketrightBig π 1 2 integraldisplay π sin 2 x dx = 0 + 1 4 bracketleftBig cos 2 x bracketrightBig π = 0 ....
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 Fall '08
 RAdin
 Trigonometry, dx, Partial fractions in integration

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