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 HW #8 / Exam #2 (including hints)

April 1, 2015 at 2:09 pm #30031
Suman GanguliParticipantExam #2 will be on Wednesday, April 15. HW #8 consists of exercises to help you prepare for the exam.
HW #8, due Wednesday, April 15:
From the Final Exam Review sheet (handed out in class, also available here http://www.citytech.cuny.edu/academics/deptsites/mathematics/docs/review/MAT1575FinalReview.pdf):
 #5 (Review Sec 7.3, Trig Substitutions, esp Examples 1,2,4)
 #6 (Review Sec 7.5, Partial Fractions, esp Examples 1,2,5)
 #7 (Review Sec 7.6, Improper Integrals)
Sec 6.1 (pp361362): #1, 3, 4, 27
Sec 6.3 (p381): #1, 3, 5, 6 This topic was modified 7 years, 11 months ago by Suman Ganguli.
April 4, 2015 at 12:11 am #30053
Suman GanguliParticipantYou may find the exercises on the Final Exam Review sheet challenging. I’ll post some hints that will hopefully help you get through them.
For #5, all of these integrals can be solved used trig substitutions. Here’s an outline of how to approach such integrals:
 Refer to the summary table at the end of Sec 7.3 to figure out which trig substitution to make (you don’t need to memorize these; I will provide a table like this on the exam)
 After you make the substitution, simplify the new “theta” integrand as much as possible. You will usually need to make use of the definitions of the trig functions, such as tan t = sin t/cos t, sec t = (1/cos t) and csc t = (1/sin t)
 Often the simplified theta integral will be one you solve by just looking it up in the table of trig integrals (in the book in Sec 6.2; again, you don’t need to memorize these, I will provide them on the exam).
But of the 4 integrals in #5, only #5(a) requires you to do that (hint: after you simplify you should have the integral of csc^2 theta, which is in the table).
For #5(b) and (c), after you make the trig substitution and simplify you are left with a relatively simple trig integrals that you can integrate via a simple usubstitution (hint: u = sin theta in both cases)
For #5(d), the theta integral simplifies to a very simple integral that you can integrate directly
 But remember that even after you solve the trig integral in terms of theta, you’re not done. You have to put the result in terms of the original variable x. Again, refer to the Sec 7.3 summary, in particular to the triangles that correspond to each of the 3 types of trig substitutions.
These triangles are necessary b/c the thetaantiderivative will usually involve trig functions, e.g., the answers for #5(b), (c) & (d) all involve sin theta. Use the triangle to figure out what sin theta is in terms of x (using sin theta = opposite/hypotense)
If these hints don’t make sense, it may help to reread the Examples in Sec 7.3 while following my outline in parallel.
 This reply was modified 7 years, 11 months ago by Suman Ganguli.
April 6, 2015 at 10:48 pm #30061
Suman GanguliParticipantHere are some hints on #6 from the Final Exam Review sheet:
These are all integrals that can be solved by the method of partial fractions. #6(b) and (c) are the simplest type of partial fractions: where the denominator factors into distinct linear factors. You should be familiar with these from the examples we did in class and from and homework exercises from that section (Sec 7.5). See also Examples 1 & 2 in that section if you need a review.
#6(a) is a little more complicated. You can factor the denominator of the integrand into x^3 + 6x = x(x^2 + 6), but note that the latter quadratic factor can’t be factored. This is what’s called an “irreducible quadratic factor”; to get a partial fractions decomposition it’s not sufficient to look for constants A and B such that
(3 – 5x^2)/x(x^2+6) = A/x + B/(x^2+6)
Instead, the numerator above the irreducible quadratic factor may be a linear function itself, i.e., we need to solve for A, B, C such that
(3 – 5x^2)/x(x^2+6) = A/x + (Bx+C)/(x^2+6)
To solve for A, B, C, the first step is the usual one: multiply though by the denominator of the LHS in order to clear the fractions:
(3 – 5x^2) = A(x^2+6) + (Bx+C)x
You can solve for A by the usual method: by plugging in the value of x which makes the “other” term 0 (in this case x=0).
But that doesn’t work for solving for B and C, since x^2+6 can’t be made to equal 0 (that’s why it’s irreducible!). Solving for B and C involves some more work: multiply out the RHS, group the x^2, x and constant terms, and now equate the coefficients of the x^2, x and constant terms (“equate the like terms”).
If this doesn’t make sense, read Sec 7.5, Example 5 in detail, and then try to finish this exercise.
 This reply was modified 7 years, 11 months ago by Suman Ganguli.
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