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No correct answer?

  1. May 18, 2014 #1

    gingermom

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    1. The problem statement, all variables and given/known data

    ∫(1+cos(x))/sin(x) dx

    This is a multiple choice with the following options

    a. Ln|1-cos(x)| +C

    b. Ln|1+cos(x)| +C

    c. sin(x) +C
    d. csc(x)+tan(x) + C
    e. csc(x) +cot(x) +C

    2. Relevant equations



    3. The attempt at a solution

    ∫(1+cos(x))/sinx dx )
    ∫(1/sin(x)+cos(x)/sin(x) )dx
    ∫(1/sin(x) dx +∫(cos(x)/sin(x) dx
    ∫(csc(x) dx +∫(cot(x) dx

    this give me the integral that is listed as the antideriviative of E

    When I differentiate all of the answers I get the following
    a. Ln|1-cos(x)| +C - sin(x)/(1-cos(x))

    b. Ln|1+cos(x)| +C - sin(x)/(1+cos(x))


    c. sin(x) +C - cos(x)
    d. csc(x)+tan(x) + C −cot(x)csc(x)+sec^2(x)
    e. csc(x) +cot(x) +C −cot(x)csc(x)−csc^2(x)

    I can't make any of those turn into (1+cos(x))/sin(x)

    So am I doing something really wrong - and if so can someone point me in the write direction, or is there an error in the optional answers?
     
  2. jcsd
  3. May 18, 2014 #2

    gingermom

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    I think I may have figured this out. substitute 1+cos(x) for U du = -sin(x) so that would be the integral of -u
     
  4. May 18, 2014 #3

    gingermom

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    If I make sin(x) = u then dx= du/cos(x)

    then I have the ∫1/sin(x)dx + ∫1/u du=∫1/sin(x)dx +ln(u) but I am stuck at the integral for 1/sin(x)
     
  5. May 18, 2014 #4

    gingermom

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    Well my calculator gives me ln(1-cos(x)) + C
     
  6. May 18, 2014 #5
    The most straightforward way to compute your integral is to rewrite the integrand;
    $$\int\frac{1+\cos x}{\sin x}\ dx=\int\frac{1}{\sin x}+\frac{\cos x}{\sin x}\ dx=\int\csc x+\frac{\cos x}{\sin x}\ dx$$
    Dealing with the integral on the right is easy enough of you know the antiderivative of ##\csc x## and not particularly easy if you don't.

    The alternative is to use a nifty trick which you wouldn't maybe think to use if you haven't seen it. The same trick is used to make the integral easier to compute and to turn one of your derivatives into one of the options. If you multiply the integrand on top and bottom by ##1-\cos x## and use some trig identities to "simplify", you might find that you have something that lends itself to a very easy ##u##-sub. Like wise, if you multiply the derivative from option (a) on top and bottom by ##1+\cos x## and simplify, you'll get something that looks more (exactly) like the integrand.

    For what it's worth, I would advise against using the advanced functions of your calculator to find answers to calculus problems. My experience is that students who do too much of that learn bad habits and don't really learn how to do the problems. I understand the desire to know what the answer is, and sometimes it can be helpful. But my experience, again, is that it's better to wait until you have an answer and then use your calculator to check it than it is to use the calculator to find the answer an then try to reverse-engineer a solution.
     
  7. May 18, 2014 #6

    gingermom

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    Thank you! - it is why I tried to figure it out first - and you are correct I likely would have never thought of multiplying by (1-cos(x))/(1-cos(x)), although I remember coming across doing something similar ( different version of the equivalent to 1) to find the anti-derivative of csc(x), so maybe since I had determined it had to be a or b I would have eventually gotten there, but probably not nd certainly not anytime soon. I will have to remember to think out of the box for trig functions that have integrals.
     
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