Math 181, Exam 2, Fall 2014 Problem 1 Solution. sin 3 (x) cos(x) dx.

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1 Math 8, Eam 2, Fall 24 Problem Solution. Integrals, Part I (Trigonometric integrals: 6 points). Evaluate the integral: sin 3 () cos() d. Solution: We begin by rewriting sin 3 () as Then, after using the substitution we have the following result: sin 3 () sin() sin 2 () sin()( cos 2 ()). u cos(), du sin() d sin 3 () cos() d sin()( cos 2 ()) cos() d, ( cos 2 ()) cos() sin() d, }{{}}{{} u 2 du }{{} u ( u 2 ) u ( du), ( u 2 )u /2 du, ( u /2 u 5/2) du, 2 3 u3/ u7/2 + C, 2 3 (cos())3/ (cos())7/2 + C.

2 Math 8, Eam 2, Fall 24 Problem 2 Solution 2. Integrals, Part II (Trigonometric substitutions: 6 points). Evaluate the following integral. Do not forget to simplify your answer completely the final answer should not contain inverse trigonometric functions! d. (4 + 2 ) 3/2 Solution: We begin with the substitition 2 tan θ, d 2 sec 2 θ dθ Using the above substitution, the integral transforms as follows: d (4 + 2 ) 3/2 (4 + (2 tan θ) 2 ) 2 3/2 sec2 θ dθ (4 + 4 tan 2 θ) 2 }{{} 3/2 sec2 θ dθ 4 sec 2 θ (4 sec 2 θ) 2 3/2 sec2 θ dθ 4 3/2 (sec 2 θ) 2 3/2 sec2 θ dθ 8 sec 3 θ 2 sec2 θ dθ 4 sec θ dθ cos θ dθ 4 The result of the above integral is (4 + 2 ) d 3/2 4 cos θ dθ 4 sin θ + C Finally, we write the answer in terms of. To do this, we return to the substitution 2 tan θ and rewrite the equation as tan θ 2 opposite adjacent where opposite is the side opposite the angle θ in the right triangle below and adjacent is the side adjacent to θ. θ 2 The hypotenuse of the right triangle is by way of Pythagoras Theorem. Thus, an epression for sin θ in terms of is sin θ opposite hypotenuse Our final answer is now (4 + 2 ) d 3/2 4 sin θ + C C. Page 2

3 Math 8, Eam 2, Fall 24 Problem 3 Solution 3. Integrals, Part III (Partial fractions decompositions: 8 points). Evaluate the integral: d. Solution: The partial fraction decomposition method will be used to evaluate the integral. The denominator factors into: ( + 3)( 2) Since the denominator is a product of distinct linear factors, the integrand may be decomposed as follows: A B 2 We find the unknown constants A and B by first clearing denominators: and then making two substitutions: A( 2) + B( + 3) 2 A(2 2) + B(2 + 3) which yields B 5 3 A( 3 2) + B( 3 + 3) which yields A 5 Substituting these values into the decomposition and integrating yields the following result: ( ) d d 2 5 ln ln 2 + C 5 Page 3

4 Math 8, Eam 2, Fall 24 Problem 4 Solution 4. Integrals, Part IV (Improper integrals: 6 points). Evaluate the following improper integral or show that it diverges. e d. Solution: The upper limit of integration makes the integral improper. Thus, we replace the upper limit with b and take the limit of the integral as b. e d lim b b e d. [ point] The integral may be evaluated using integration by parts. We make the following definitions: u, dv e d which yield Thus, using the integration by parts formula, we have b The value of the improper integral is then b du d, v e. u dv uv b b v du e d ( e ) b e d lim b e b + b b e e b e ( + ) b + b e b + e b + where the limit of b + e b as b is since b + e b for large b. b ( e ) d e d e d lim b + b }{{ e b + } Page 4

5 Math 8, Eam 2, Fall 24 Problem 5 Solution 5. Integrals, Part V (Numerical integration: 8 points). Evaluate the integral using: 2 (2 + ) 2 d (a) analytical methods, to obtain the eact solution; (b) the Midpoint Method for numeral integration, with n 2 subintervals. Simplify your answer! Compare the two results by computing the absolute and the relative errors. Solution: (a) The Fundamental Theorem of Calculus yields the following eact solution: 2 (b) The width of each subinterval of [, 2] is given by [ (2 + ) 2 d ] The two subintervals of [, 2] are [, ] and [, 2]. The corresponding midpoints of these subintervals are The value of M 2 is then m 2, m M 2 [f(m + f(m 2 )] [ ( ) ( )] 3 f + f 2 2 ( )2 ( ) Page 5

6 Math 8, Eam 2, Fall 24 Problem 6 Solution 6. Sequences (5 points). For each of the following sequences, determine whether they have a limit or not, and if yes, then compute the limit. (a) a n 2n+ n 2 2 for all n ; (b) b n sin(nπ) for all n. Solution: (a) The limit of the sequence is (b) The terms of the sequence are lim a n lim n n Thus, the limit of the sequence is. 2n + n 2 2 n 2 n 2 lim n 2 n + n 2 2 n 2 + {a n } {sin(π), sin(2π), sin(3π),..., sin(nπ),...} {,,,...,,...} Page 6

7 Math 8, Eam, Fall 24 Problem 7 Solution 7. Integrals, Part VI (7 points). (a) Decompose the polynomial q() into a product of irreducible factors, noticing that q(). (b) Decompose the rational function r() (c) Evaluate r() d. into its partial fraction decomposition. Solution: (a) We can factor q() as follows: q() ( 3 2 ) + (2 2) 2 ( ) + 2( ) ( 2 + 2)( ) where is irreducible because its discriminant, b 2 4ac 2 4()(2) 8, is negative. (b) The partial fraction decomposition of r() is Clearing denominators gives us A + B + C A( 2 + 2) + (B + C)( ) Epanding the right hand side and collecting like terms results in the equation + 2 (A + B) 2 + (C B) + (2A C) (A + B) 2 + (C B) + (2A C) Equating coefficients of n on both sides of the above equation gives us the following system of equations 2 : A + B, : C B, : 2A C 2. The solution to the system is C, B, and A. (c) The integral of r() is then ( r() d ) 2 d + 2 d d In the second integral on the right hand side above, we let u Then du 2 d 2 du d and we have d 2 u du 2 ln u 2 ln(2 + 2) Thus, we have r() d d d ln 2 ln(2 + 2) + C Page 7

8 Math 8, Eam 2, Fall 24 Problem 8 Solution 8. Integrals, Part VII (6 points). Evaluate the following integral: d. Solution: Let 2 sec θ and d 2 sec θ tan θ dθ. Then d (2 sec θ) 2 (2 sec θ tan θ dθ) (2 sec θ) sec θ tan θ 4 sec 2 θ 4 sec 2 θ 4 dθ tan θ 2 sec θ 4(sec 2 θ ) dθ tan θ 2 sec θ 4 sec 2 θ dθ tan θ 2 sec θ 2 tan 2 θ dθ tan θ 4 sec θ tan θ dθ cos θ dθ 4 4 sin θ + C Since 2 sec θ we know that sec θ 2 which means that cos θ 2. We set up the following right triangle θ 2 Using Pythagoras Theorem, the side opposite θ is 2 4. Therefore, d 4 sin θ + C C Page 8