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Homework Help: Integration by trig substitution

  1. Jul 14, 2010 #1
    1. The problem statement, all variables and given/known data
    the problem is INTEGRAL 6dz/(z^2(sqrt(z^2+9))


    z^2+ a^2 , then z=a*tan@ where a here is 3, because 3^2=9 ,
    i use @ here to represent theta

    substituting this for z;

    Int[6/(z^2(sqrt(z^2+9))]dz= Int[(6*3sec^2@d@)/(9tan^2@(sqrt(9tan^2@+9))]=
    Int[(18sec^2@d@)/(9tan^2@(sqrt(9(tan^2@+1))]=
    Int[(18sec^2@d@)/(9tan^2@(SQRT(9sec^2@))
    =2/3Int[sec^2@/(tan^2@sec@)]d@
    2/3int[sec@/tan^2@] d@
    using sec@=1/cos@ and tan@=cos@/sin@
    =2/3int[1/cos @*cos^2 @/sin^2 @] d@
    =2/3int[cos @/sin^2 @] d@...from here i dont know what to do..please healp
     
  2. jcsd
  3. Jul 14, 2010 #2

    Mark44

    Staff: Mentor

    You've done all the hard work - the rest is easy. Let u = sin@, du = cos@d@.

    Don't forget to undo both your substitutions to get back in terms of z.
     
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