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Integration problem

  1. Feb 3, 2017 #1
    1. The problem statement, all variables and given/known data
    given ## tan 2θ-tan θ≡ tan θ sec 2θ## show that ##∫ tan θ sec θ dθ = (1/2 )ln (3/2)## limits are from θ= 0 to θ=π/6

    2. Relevant equations


    3. The attempt at a solution
    ##∫ tan 2θ-tan θ dθ ##
    -(1/2 )ln cos 2θ + ln cos θ
    → ##-1/2 ln 1/2 + ln √3/2##
    ##= ln (√3)/2- (1/2)ln (1/2)##
    ok am i correct up to this point?
     
    Last edited: Feb 3, 2017
  2. jcsd
  3. Feb 3, 2017 #2
    anyone on this.................
     
  4. Feb 3, 2017 #3

    haruspex

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    That should be sec 2θ, right?
    a little careless with the parentheses.
     
  5. Feb 3, 2017 #4
    Yes sir sec2¤
     
  6. Feb 3, 2017 #5

    haruspex

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    Ok. What about my other comment?
     
  7. Feb 3, 2017 #6
    ##ln \frac{3^1/2} 2## -##\frac1 2## ##ln\frac 1 2##
     
    Last edited: Feb 3, 2017
  8. Feb 3, 2017 #7
    is that better? i have a problem using latex
     
  9. Feb 3, 2017 #8
    is this correct?
     
    Last edited: Feb 3, 2017
  10. Feb 3, 2017 #9

    haruspex

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    Yes, but all you needed to do was to correct from this:
    ##= ln (√3)/2- (1/2)ln (1/2)##
    to
    ##= ln ((√3)/2)- (1/2)ln (1/2)##
    so as to distinguish it from
    ##= (ln (√3))/2- (1/2)ln (1/2)##
     
  11. Feb 4, 2017 #10
    now how is this equal to ## \frac 1 2## ln##\frac 3 2##?
     
  12. Feb 4, 2017 #11

    haruspex

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    Just a bit of manipulation. How else could you write ln(√3)?
     
  13. Feb 4, 2017 #12
    i am getting ##\frac 1 2## ln 3-ln 2 - (##\frac 1 2## ln 1 - ##\frac 1 2##ln 2)
    which gives me ##\frac1 2## ln 3- ln 2 - ##\frac 1 2##ln 1 +##\frac 1 2## ln2
    this becomes ##\frac 1 2##ln 3 + ##\frac 1 2##ln 2 - ln 2
    this becomes
    ##\frac 12##ln 3 +ln##\frac {2^1/2} 2##
    ##\frac 1 2##ln 3 + ln ##{2^-1/2}##
    ##\frac 1 2##ln 3 - ##\frac 1 2## ln 2
    ##\frac 1 2##ln ##\frac 3 2##

    is this now correct? greetings from Africa............
     
    Last edited: Feb 4, 2017
  14. Feb 4, 2017 #13

    haruspex

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    Yes, but you can get there a little faster.
    ##\frac 1 2##ln 2 - ln 2=##-\frac 1 2##ln 2
     
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