EXAM. Practice for Second Exam. Math , Fall Nov 4, 2003 ANSWERS

Transcription

1 EXAM Practice for Second Eam Math , Fall 003 Nov 4, 003 ANSWERS

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3 Problem 1. In each part, find the integral. A. d (4 ) 3/ Make the substitution sin(θ). d cos(θ) dθ. We also have Then, we have d/dθ cos(θ), so so sin (θ) 4[1 sin (θ)] 4 cos (θ), (4 ) 3/ 4 3/ [cos (θ)] 3/ 8 cos 3 (θ). Plugging into the integral gives us d (4 ) 3/ 4 sin (θ) 8 cos 3 cos(θ) dθ (θ) sin (θ) cos (θ) dθ tan (θ) dθ tan(θ) 1 dθ (Reduction Formula (3)) (*) tan(θ) θ + C. From above, we have / sin(θ), so θ sin 1 (/). We also have cos(θ) 4 /, so Plugging into (* ), we get tan(θ) sin(θ) cos th 4 4. (4 ) 3/ d 4 sin 1 (/) + C B. cos() d. 1

4 Use integration by parts. We want to differentiate the factor to get rid of it. Thus, in the integration by parts formula (1) uv d uv uv d we set u and v cos(). We then have u 1 and v v d cos() d 1 sin() (why?) Plugging into (1) gives us cos() d 1 sin() [ ] 1 (1) sin() d 1 sin() 1 sin() d 1 sin() 1 [ 1 ] cos() + C 1 sin() + 1 cos() + C. 4 C. 3 1 d If you write the integral as 1 d, you may see that the simple substitution () u 1 will work. This substitution gives us du/d and so d ( 1/) du. From () we have 1 u. Plugging into the integral gives us 3 1 d (1 u) ( u 1 ) du 1 (u 1/ u 3/ ) du 1 [ 3 u3/ ] 5 u5/3 + C 1 3 (1 ) 3/ (1 ) 5/ + C

5 D. ln() d Use integration by parts (Formula (1)) with u ln() and so v 1. Then, u 1/ and v v d 1 d. Plugging into the integration by parts formula(1) gives [ ] 1 ln() d ln() () d ln() 1 d ln() + C. E. ln() d If we use integration by parts and differentiate ln(), all that will be left is powers of. Thus, set u ln() and v. We then have u 1/ and v v d d 3 3. Plugging into the integration by parts formula (1) we have [ ][ ] ln() d ln() d ln() 1 d ln() 1 [ ] 3 + C ln() C 3

6 F. 4 + d. Make the trigonometric substitution (3) tan(θ). Then we have We also have d dθ sec (θ) d sec (θ) dθ tan (θ) 4[1 + tan (θ)] 4 sec (θ) and so (4) sec (θ) sec(θ). Plugging into the integral we get 4 + d sec(θ) sec (θ) dθ 4 sec 3 (θ) dθ [ 1 4 sec(θ) tan(θ) + 1 ] sec(θ) dθ (Reduction Formula (5)) sec θ tan(θ) + ln sec θ + tan(θ) + C ln C, (From (3) and (4)) ln C ln C, where in the last (optional) step, we ve absorbed ln(1/) into the arbitary constant C. G. 1 d 4

7 Note that the integrand is defined for 1 and 1. To handle the case 1, make the trig substitution sec(θ). We have and we have d dθ sec(θ) tan(θ) d sec(θ) tan(θ) dθ 1 sec (θ) 1 tan (θ) and so we have 1 tan(θ). Plugging into the integral gives 1 tan(θ) d sec(θ) tan(θ) dθ sec(θ) tan (θ) dθ tan(θ) θ + C, (From Reduction Formula (3) ) 1 sec 1 () + C Thus, we have 1 (5) d 1 sec 1 () + C, 1. Now consider the case 1. To deal with this case make the substitution u (so u 1) in the integral. Of course d du and we have 1 ( u) 1 d ( 1) du u u 1 du u u 1 sec 1 (u) + C, (from (5)) ( ) 1 sec 1 ( ) + C 1 sec 1 ( ) + C Combining the two case, we have the formula { 1 1 sec 1 () + C, 1 d 1 sec 1 ( ) + C, 1. We can easily combine the two cases into one, using absolute value. Thus, our final answer is 1 d 1 sec 1 + C. 5

8 H d A quick check with the quadratic formula shows that doesn t factor. Thus, we want to complete the square. That is, we want to write + + ( h) + k h + h + k Comparing coefficents shows that h so h 1 and h + k 1 + k so k 1. Thus, we have + + ( + 1) + 1. Thus, our integral becomes ( + 1) ) d. In this integral, substitute u + 1. With the substitution, d du and u 1. Thus, our integral becomes 1 (u 1) ( + 1) ) d 1 + u du u 1 + u du u 1 + u du u du 1 ln(1 + du u ), (guess and correct method) 1 + u 1 ln(1 + u ) tan 1 (u) + C 1 ln[1 + ( + 1) ] tan 1 ( + 1) + C 1 ln( + + ) tan 1 ( + 1) + C. Problem. In each part, give the form of the partial fraction decomposition. This is a formula involving undetermined coefficients. Do not find the coefficients! (No calculation is required). A ( 1)( )( + 3) ( 1)( )( + 3) A 1 + B + C

9 B. 1 ( + 1) 1 ( + 1) A + B + C + 1. C ( + 1) ( + 1) A + B + C D + E ( + 1). D. 3 ( ) ( + ) ( 1) 3 ( ) ( + ) ( 1) A + B ( ) + C + + D ( + ) + E 1. Problem A. 3. In each part, find the partial fraction decomposition of the given rational function (i.e., find the coefficients). 5 1 ( 1)( + 1) The form of the partial fraction decomposition should be (6) 5 1 ( 1)( + 1) A + B 1 + C + 1. Clearing the denominators in this equation leads to the equation (7) 5 1 A( 1)( + 1) + B( + 1) + C( 1). 7

10 Setting 0 in this equation yields the equation 1 A( 1)(1) + B(0)(1) + C(0)( 1) which gives us 1 A, so A 1. Setting 1 in equation (7) yields 4 B(1)(), so B. Finally, setting 1 in equation (7) yields the equation 6 C, so C 3. Plugging these values for A, B and C into equation (6) gives the answer 5 1 ( 1)( + 1) B. + 1 ( 1) The form of the partial fraction decomposition is (8) + 1 ( 1) A + B + C 1. Clearing denominators in this equation gives (9) + 1 A( 1) + B( 1) + C. Epanding this out gives + 1 A A + B B + C (A + C) + ( A + B) B. Equating coefficents gives the system of equations A + C 1 A + B 0 B 1 These are easy to solve, giving A 1, B 1 and C. Plugging these values into (8) gives + 1 ( 1)

11 C ( + 1) The form of the partial fraction decomposition is (10) ( + 1) A + B + C + D + 1. Clearing the denominators gives A( + 1) + B( + 1) + (C + D) A 3 + A + B + B + C 3 + D (A + C) 3 + (B + D) + A + B. Equating coefficents leads to the following system of equations A + C B + D A 0 B 1. These are easy to solve and the solution is A 0, B 1, C and D 3. Plugging these values into equation (10) gives ( + 1)