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Simple integral

  1. Jun 15, 2017 #1
    1. The problem statement, all variables and given/known data

    Integrate: cos2x/[cos^2 (x).sin^2 (x)]
    2. Relevant equations

    ▪cos2x=1-2sin^2 (x)
    ▪2sinxcosx=sin2x
    ▪1/sinx = cosecx
    ▪Integration of cosec^2 (ax+b)=[-cot(ax+b)]/a
    ▪Integration of sec^2 (x)=tanx
    3. The attempt at a solution
    I have attached my solution,but the answer is not matching with the correct answer (written in the last line).I wrote the given answer as well coz it may be a manipulation of my answer,which i can't see(doubtful,but not impossible).If anyone could just point out the line where I'm going wrong..
     

    Attached Files:

  2. jcsd
  3. Jun 15, 2017 #2

    BvU

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    An easy check is to differentiate your result !
     
  4. Jun 15, 2017 #3

    BvU

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    And you could convert ##\cot 2x ## to ##\ \ \displaystyle {\cot^2 x -1 \over 2\cot x} ## :smile:
     
  5. Jun 15, 2017 #4
    Of course,why didn't I think of that?
    Thanks!!
     
  6. Jun 15, 2017 #5

    Ray Vickson

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    You are developing a bad habit, which you should stop right away if you want to continue posting to PF. Most helpers will not look at images of handwritten solutions; I, for one, will not. You may be lucky to find somebody willing to help by looking at your images, but please do not keep doing it; the PF standard is to type out your work, and it really is not very difficult. For example, you can write ##\int_a^b x/(x^2+a^2) \, dx## in plain text as int{ x/(x^2+a^2) dx, x=a..b} (or as int_{x=a..b} {x/(x^2+a^2) dx}) and that is perfectly readable. Just be careful to use parentheses, so that ##\frac{a + b}{c}## is written as (a+b)/c, NOT as a + b/c (which means ##a + \frac{b}{c}##).

    Please try to reserve images for things like drawings, diagrams and/or data tables.
     
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