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Beginning Trignometric Antiderivative

  1. Jan 19, 2009 #1
    1. The problem statement, all variables and given/known data

    Okay, I think I'm finally getting the hang of these antiderivatives. However, I'm still stumbling some on trigonometric functions.

    Find the antiderivative of [tex]f(\theta) \ = \ \frac{1 + \cos^2{\theta}}{\cos^2{\theta}} [/tex]


    2. Relevant equations


    [tex] f(\theta) \ = \ \frac{1 + \cos^2{\theta}}{\cos^2{\theta}} [/tex]



    3. The attempt at a solution

    1.) [tex] f(\theta) \ = \ \frac{1 + \cos^2{\theta}}{\cos^2{\theta}} [/tex]

    2.) [tex] f(\theta) \ = \ \frac{1}{\cos^2{\theta}}} \ + \ \frac{cos^2{\theta}}{cos^2{\theta}} [/tex]

    3.) [tex] f(\theta) \ = \ \frac{1}{\cos^2{\theta}}} \ + \ 1 [/tex]

    4.) Trigonometric Identity: [tex] \frac{1}{\cos^2{\theta}}} \ = \ \csc^2{\theta} [/tex]

    5.) [tex] f(\theta) \ = \ \csc^2{\theta}\ + \ 1 [/tex]

    6.) [tex] F(\theta) \ = \ -\cot{\theta} \ + \ \theta [/tex]


    Where have I gone wrong? I know the answer isn't correct, but I'm not sure what I have done wrong here?
     
    Last edited: Jan 19, 2009
  2. jcsd
  3. Jan 19, 2009 #2

    Dick

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    Homework Helper

    1/cos(t)=sec(t). Not csc(t). And you probably don't want to write the antiderivative of F as F'. Call it G or something.
     
  4. Jan 19, 2009 #3
    Ah, crap! I don't know where I got [tex] \frac{1}{\cos^2{\theta}}} \ = \ \csc^2{\theta} [/tex] from. That's obviously not right. Thank you for pointing out my stupid mistake.

    So....

    1.) [tex] f(\theta) \ = \ \frac{1 + \cos^2{\theta}}{\cos^2{\theta}} [/tex]

    2.) [tex] f(\theta) \ = \ \frac{1}{\cos^2{\theta}}} \ + \ \frac{cos^2{\theta}}{cos^2{\theta}} [/tex]

    3.) [tex] f(\theta) \ = \ \frac{1}{\cos^2{\theta}}} \ + \ 1 [/tex]

    4.) Trigonometric Identity: [tex] \frac{1}{\cos^2{\theta}}} \ = \ \sec^2{\theta} [/tex]

    5.) [tex] f(\theta) \ = \ \sec^2{\theta}\ + \ 1 [/tex]

    6.) [tex] F(\theta) \ = \ \tan{\theta} \ + \ \theta [/tex]


    Now, would it be [tex] F(\theta) \ = \ \tan{\theta} \ + \ \theta \ + \ C [/tex] or just [tex] F(\theta) \ = \ \tan{\theta} \ + \ \theta [/tex] ?
     
  5. Jan 19, 2009 #4

    jgens

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    Gold Member

    You need to include the constant C. While it may seem trivial or non-important, you can actually contrive anti-derivatives, which, when simplified yield 1 = 0 (or some other non-sensical equality) if you neglect the constant.
     
    Last edited: Jan 19, 2009
  6. Jan 20, 2009 #5
    Great. Thanks to everyone for their help!
     
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