MATH QUIZ 3 1/2. sin 1 xdx. π/2. cos 2 (x)dx. x 3 4x 10 x 2 x 6 dx.

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1 NAME: I.D.: MATH 56 - QUIZ 3 Instruction: Each problem is worth of point in this take home project. Circle your answers and show all your work CLEARLY. Use additional paper if needed. Solutions with answer only and without supporting procedures will have little credit. In some of the applications of integration by parts, you may need to apply the integration by parts more than once, (such as Eercise 7 on Page 39). You need to show all these by-parts application for a full credit. (If you have consulted with the Math Learning Center, please indicate which problem you get help from MLC).. (Eercise 7, Pager 39) Compute sin()d.. (Eercise, Pager 39) Compute 5 ln()d. 3. (Eercise, Page 39) Compute 4. (Eercise 5, Page 39) Compute 5. (Eercise 5, Page 39) Compute / π/ sin d. cos ()d. tan 3 () sec()d. Instruction: For Problems 6-8, first write down what the correct substitution is and perform the corresponding computation for the substitution. d 6. (Eercise 45, Page 39) Compute (Eercise 53, Page 39) Compute 9 d. 8. (Eercise 43, Page 39) Compute 5 d (Eercise 9, Page 37) Compute ( )( + 5) d.. (Eercise 6, Page 37) Compute d. Questionnaires: In order for me to provide a better teaching approach to help you learn the material, I need to know your preparedness for this course. Please answer the following questionnaires: (A) In your Calculus class, did you learn how to compute an antiderivative of a function? (Circle one: Yes No I am not sure) (B) In your Calculus class, did you learn how to evaluate a definite integral? (Circle one: Yes No I am not sure) You can write down any comments if you consider it difficult to answer these questionnaires.

2 . (Eercise 7, Pager 39) Compute Solutions: sin()d. Solutions: We use integration by parts twice. sin()d = cos() + cos() = cos() + sin() By-Parts: u =, dv = sin()d sin() By-Parts: u =, dv = cos()d = cos() + sin() + cos() + C. (Eercise, Pager 39) Compute 5 ln()d. Solutions: We use integration by parts. 5 ln() = 6 6 ln() 6 6 d = ln() 6 = (Eercise, Page 39) Compute ln() 6 36 d + C / By-Parts: u = ln(), dv = 5 d d simplifying the integrand sin d. Solutions: We use integration by parts. In this eercise, we have no choice other than u = sin () and dv = d. / sin d = sin () d By-Parts: u = sin (), dv = d = sin ( ) = π + w = π (Eercise 5, Page 39) Compute Solutions: π/ w dw substitution: w =, dw = d sin(π/6) = / and so sin (/) = π/6 cos ()d. Need to use half angle formula cos () = + cos(). π/ π/ cos ()d = ( + cos())d = ( + sin() ) π/ = ( π + sin(π) ) = π (Eercise 5, Page 39) Compute tan 3 () sec()d. Solutions: The differentiation formulae (tan()) = sec () and (sec()) = sec() tan()

3 suggest that we use substitution u = sec() and du = sec() tan()d. As tan 3 () sec() = tan () tan() sec() = (sec () ) tan() sec(), we have tan 3 () sec()d = = (sec () ) tan() sec()d (u )du substitution: u = sec(), du = tan() sec()d = u3 3 u + C = sec3 () sec() + C 3 Instruction: For Problems 6-8, first write down what the correct substitution is and perform the corresponding computation for the substitution. d 6. (Eercise 45, Page 39) Compute + 6. Solutions: Among the quantities + 6, and 4, + 6 is the largest and should be the hypotenuse of the right triangle. Pick an angle θ so that tan θ = 4, or = 4 tan θ. Hence d = 4 sec (θ)dθ. Compute + 6 = 6 tan () + 6 = 4 sec(θ). d (Eercise 53, Page 39) Compute = = 4 sec (θ)dθ 4 sec(θ) sec(θ)dθ = ln sec(θ) + tan(θ) + C = ln C = ln ln(4) + C cancellation of common factors = ln C choose C = C ln 4 9 d. Solutions: Among the quantities 9, and 3, is the largest and should be the hypotenuse of the right triangle. Pick an angle θ so that sec θ = 3, or = 3 sec θ. Hence d = 3 sec(θ) tan(θ)dθ. Compute 9 = 9 sec () 9 = 3 tan(θ). 3 sec(θ) 9 d = 3 sec(θ) tan(θ)dθ 3 tan(θ) = 3 sec (θ)dθ cancellation of common factors = 3 tan(θ) + C = 9 + C 8. (Eercise 43, Page 39) Compute 5 d. Solutions: Among the quantities 5, and 5, 5 is the largest and should be the hypotenuse of the right triangle. Pick an angle θ so that sin θ = 5, or = 5 sin θ. Hence d = 5 cos(θ)dθ. Compute 5 = 5 5 sin () = 5 cos(θ). 5 d = 5 sin θ 5 cos(θ) 5 cos(θ)dθ

4 = 5 sin dθ cancellation of common factors θ = csc (θ)dθ Identity: csc(θ) sin(θ) = 5 = 5 cot(θ) + C = cos(θ) 5 sin(θ) + C 5 = + C use the right triangle to see cos(θ) = (Eercise 9, Page 37) Compute ( )( + 5) d Solutions: Note that in the fraction, the degree of the numerator is less than that of the denominator. Hence we can proceed the partial fractions. Let A and B be unknowns such that 9 ( )( + 5) = A + B + 5 A( + 5) + B( ) =. ( )( + 5) Method : Now we 9 = A( + 5) + B( ). Let = 5. Then 5 9 = B( 5 ), or B =. Let =, then 9 = A( + 5), and so A =. Method : Epand the product and collect like terms to get 9 = A + 5A + B B = (A + B) + (5A B). Compare the coefficients to see that = A + B and 9 = 5A B. From A + B = we have A = B. Substitute A = B into 9 = 5A B to get 9 = 5( B) B = 5 7B. This gives B =. As = A + B = A +, we have A =. 9 Now using ( )( + 5) = + to get d ( )( + 5) d = + d = ln + ln C (Eercise 6, Page 37) Compute 6 d. Solutions: Note that in the fraction, the degree of the numerator is not less than that of the denominator. Hence we can not proceed the partial fractions. Instead, we need to do division first. Hence 3 4 = ( 6)( + ) or = ( 4 6 d = ) d = d. Now the degree of 3 4 is less than that of 6. We proceed by factoring 6 = ( + )( 3), and try to find the partial fractions corresponding to the integrand Let A and B be unknowns such that 3 4 ( + )( 3) = A + + B 3 A( 3) + B( + ) =. ( + )( 3)

5 Method : Now we 3 4 = A( 3) + B( + ). Let = 3. Then 9 4 = B(3 + ), or B =. Let =, then 6 4 = A( 3), and so A =. Method : Epand the product and collect like terms to get 3 4 = A 3A + B + B = (A + B) + (B 3A). Compare the coefficients to see that 3 = A + B and 4 = B 3A. From A + B = 3 we have A = 3 B. Substitute A = 3 B into 4 = B 3A to get 4 = B 3(3 B) = 5B 9. This gives B =. As 3 = A + B = A +, we have A = 3 =. 3 4 Now using ( + )( 3) = + + to get 3 3 ( 4 6 d = ) d 6 = d = + + d + + d 3 = + + ln + + ln 3 + C.

6 Comments on solutions of students: I will comment on the errors we made in this group of eercises. You make learn from the others errors also.. (Eercise 7, Pager 39) Compute sin()d. Many chose u = sin() and dv = d. They might have forgotten what was discussed in class. This choice will make the integration more complicated than the original, and so should be rejected. Let us repeat the discussionin class again: In general, to apply integration by parts on integrals like p() sin()d (or p() cos()d) for a polynomial p(), always set u = p() and dv = sin()d or v = sin()d = cos(). This choice is successful as du will lower the degree of p(), and eventually vanish the polynomial in successive applications of the by-parts technique. Another problem observed in this problem is when some students set (wrongly) dv = d, and then computed v =, instead of finding antiderivative v = dv = ()d =.. (Eercise, Pager 39) Compute 5 ln()d. The pro and con of whether we should use dv = ln()d is discussed in class. Still quite a few of us chose dv = ln()d, and messed up from there. Consequently, the rest of the computation/solution is totally wrong. Let us repeat the discussion in class again: In general, to apply integration by parts on integrals like p() ln()d for a polynomial p(), always set u = ln() and dv = p()d or v = p()d. This would be successful as the net step of du = d will turn the integrand a power function, which should be a piece of cake for majority of us who knows Calculus well. In doing Problem, several made the following mistake: Let dv = ln()d and so v =. I consider this is a differentiation error as if v = dv, then we should have d =, instead of ln(). Among those who made the correct choice, 6 students had the following error in applying chain rule: u = ln() and du d = 4. This indicates again that we need to seriously review the Calculus materials. 3. (Eercise, Page 39) Compute / sin d. As discussed in class, many chose the right u = sin (). However, the most common errors are in the differentiation of sin (). This indicates that many of us need to review and practice differentiation skill learned in Calculus. 4. (Eercise 5, Page 39) Compute π/ cos ()d. Those who messed up seen to be ignoring the discussion in class. We can repeat here again: when k is a positive even integer, we always apply the half-angle formula to handle sin k ()d and cos k ()d. Another error found is the misunderstanding of the integration bound change. See the following erroneous step: π/ cos ()d = π ( + cos())d. A few of us made this error. Since we did not change the variable, we should not change the

7 integration bounds as these bounds are for. More students made the following error: (You can find the error yourself. Had we do it carefully, this could have been avoided.) cos()d = sin() + C. 5. (Eercise 5, Page 39) Compute tan 3 () sec()d. If you know Calculus differentiation formulae well, you will see the the related differentiations are (sec()) = sec() tan() and (tan()) = sec (). The first would suggest us that sec() tan()d = d sec(). Hence the substitution u = sec() seems promising. Do the corresponding computation du = sec() tan()d, which then lead us to the correct solution. A common differentiation error: five students computed (sec()) = tan(). General remark for Problems 6-8: Some of use choose to apply a formula from the table on the back of tet, or from a reference guide, which can lead to a correct answer. But such solutions would not let us to receive the correct training of mastering the skills. 6. (Eercise 45, Page 39) Compute d + 6. A common error is that several used u = + 6, which is a reasonable trial for the uneperienced. However, when we compute du = d, and the integrand does not offer an d, we should recognize that this is not the right way. 7. (Eercise 53, Page 39) Compute 9 d. As suggested in class, it is OK if we use u = 9. Several students figured out the correct substitution = 3 sec(), and got the correct middle step solution 3 tan(θ), but they do not know how to go back to an epression in. This is because they did not do the associate computation: 9 = 9 sec () 9 = 3 tan(θ), which is a necessary step as one has to convert 9 into the corresponding trigonometry functions. 8. (Eercise 43, Page 39) Compute 5 d. The common errors here will be that when we make a substitution = 5 sin(theta), we did not compute d in terms of θ, and simply replace d by dθ, causing the error solution of the problem. Another common error is in Calculus anti-differentiation: several computed this following: sin θ dθ = ln sin θ + C. The correct way to compute the antiderivative should be sin θ dθ = csc θdθ (Eercise 9, Page 37) Compute ( )( + 5) d. Those who have errors are the ones who did not look at their notes.. (Eercise 6, Page 37) Compute d.

8 Some of us did not check the degree requirement and go ahead to find the corresponding partial fraction without first doing division. Most of the common errors are in the algebra when performing division.