Integration by trig substitution

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SUMMARY

The forum discussion centers on solving the integral ∫(6dz/(z²√(z²+9))). The user employs trigonometric substitution by letting z = 3tan(θ), leading to a transformed integral involving secant and tangent functions. The discussion highlights the simplification process, ultimately guiding towards the substitution u = sin(θ) for further evaluation. The final step emphasizes the importance of reverting back to the original variable z after solving the integral.

PREREQUISITES
  • Understanding of trigonometric identities, specifically secant and tangent functions.
  • Familiarity with integration techniques, particularly trigonometric substitution.
  • Knowledge of basic calculus concepts, including integrals and substitutions.
  • Ability to manipulate algebraic expressions involving square roots.
NEXT STEPS
  • Study advanced integration techniques, focusing on trigonometric substitution methods.
  • Learn about the properties and applications of secant and tangent functions in calculus.
  • Explore the process of reverting substitutions in integrals to original variables.
  • Practice solving integrals involving square roots and trigonometric functions for proficiency.
USEFUL FOR

Students studying calculus, particularly those tackling integration problems, and educators seeking to enhance their teaching methods in trigonometric substitution techniques.

doothang
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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 don't know what to do..please healp
 
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doothang said:

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 don't 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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