web-7.1 - Web 7.1 In w1–w4, integrate by parts. w1. x3 ex...

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Unformatted text preview: Web 7.1 In w1–w4, integrate by parts. w1. x3 ex dx = 1 x2 e− 2 ”− ”− 1 2 2 [x2 ] ex x dx 1 x2 e 2x dx by parts 2 ex 2x dx by B3 2 2 = x2 = = 1 x2 e +C 2 w2. Do 35 in Ex 7.1 in the book without first substituting. Solution. √ 2 2 √ π θ cos θ dθ = √ π [θ ] (cos θ θ dθ) 2 2 π √ 3 2 π = 1 θ2 sin θ2 − 2 π 1 sin θ2 2θ dθ √ = · · · π 2 2 √ w3. Do 30 in Ex 7.1 in the book. Solution. 1 0 √ r3 dr = 4 + r2 1 1 0 [r2 ] (4 + r2 )− 2 r dr (4 + r2 ) 2 2r dr 1 1 1 = r2 (4 + r2 ) 2 − w4. x3 dx √ = 5 9 + x2 1 0 = ··· [x2 ] (9 + x2 )− 5 x dx 4 5 (9 + x2 ) 5 2x dx 8 = x2 = 4 15 (9 + x2 ) 5 − 24 9 4 25 52 (9 + x2 ) 5 + C x (9 + x2 ) 5 − 72 8 In w5 and w6, integrate by parts twice. Then transpose the desired unknown integral from the right side to the left side of the equation. Then solve the equation for this unknown integral. w5. [eax ] (cos bx dx) = eax 1 a sin bx − b b [eax ] (sin bx dx) 1 a ax − cos bx = eax sin bx − e − b b b 1 a a2 = eax sin bx + 2 eax cos bx − 2 b b b − cos bx ax e a dx b eax cos bx dx Now we write the integral at the beginning of this ”string” equal to the expression after the last ”equal sign” and we get: eax a ax a2 e cos bx dx = sin bx + 2 e cos bx − 2 b b b ax eax cos bx dx Now we transpose the last term on the right over to the left and get: a2 +1 b2 eax cos bx dx = eax a sin bx + 2 eax cos bx b b Simplifying both sides of the last equation, we get: a2 + b 2 b2 eax cos bx dx = eax (b sin bx + a cos bx) b2 Multiplying both sides by b2 and then dividing by a2 + b2 , we get: eax cos bx dx = w6. Show that: eax sin bx dx = eax (a sin bx − b cos bx) + C a2 + b 2 eax (a cos bx + b sin bx) + C a2 + b 2 w7. Use the formulas in w5 and w6 to do 17 and 18 in Ex 7.1 in the book. ...
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This note was uploaded on 05/17/2011 for the course MAC 2312 taught by Professor Bonner during the Spring '08 term at University of Florida.

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web-7.1 - Web 7.1 In w1–w4, integrate by parts. w1. x3 ex...

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