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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

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# Homework Help: Integration by trig substitution

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