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Trig integration

  1. Jul 2, 2009 #1
    1. The problem statement, all variables and given/known data

    I(tan^5x,x)=I(tan^2xtan^3x,x)=I((sec^2x-1)tan^3x,x)
    =I(sec^2xtan^3x,x)-I(tan^3x,x)
    u=tanx du=sec^2x
    =I(u^3,u)-I(tan^3x,x)
    =u^4/4-I(tan^3x,x)
    =tan^4x/4-I(tan^2xtanx,x)
    =tan^4x/4-I((sec^2x-1)tanx,x)
    =tan^4x/4-I(tanxsec^2x-tanx)
    =tan^4x/4-I(tanxsec^2x,x)-I(tanx,x)
    y=tanx, dy=sec^2x
    =tan^4x/4-I(u,u)-I(tanx,x)
    =tan^4x/4-u^2/2-ln|secx|+c
    tan^4x/4-tan^2x/2-ln|secx|+c


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 2, 2009 #2
    You have a sign error between lines 8 and 9
     
    Last edited: Jul 2, 2009
  4. Jul 3, 2009 #3
    I(tan^5x,x)=I(tan^2xtan^3x,x)=I((sec^2x-1)tan^3x,x)
    =I(sec^2xtan^3x,x)-I(tan^3x,x)
    u=tanx du=sec^2x
    =I(u^3,u)-I(tan^3x,x)
    =u^4/4-I(tan^3x,x)
    =tan^4x/4-I(tan^2xtanx,x)
    =tan^4x/4-I((sec^2x-1)tanx,x)
    =tan^4x/4-I(tanxsec^2x-tanx)
    =tan^4x/4-I(tanxsec^2x,x)+I(tanx,x)
    y=tanx, dy=sec^2x
    =tan^4x/4-I(y,y)+I(tanx,x)
    =tan^4x/4-y^2/2+ln|secx|+c
    tan^4x/4-tan^2x/2+ln|secx|+c

    my texts solution is 1/4sec^4x-sec^2x+ln|secx|+c
     
  5. Jul 3, 2009 #4

    Cyosis

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    Homework Helper

    To obtain your text's solution you can simply do what you've been doing all along. That is use the identity [itex]\tan^2x=\sec^2x-1[/itex].
     
    Last edited: Jul 3, 2009
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