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Integration by trig substitution

  • Thread starter doothang
  • Start date
  • #1
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Homework Statement


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
 

Answers and Replies

  • #2
33,632
5,287

Homework Statement


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
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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