Evaluating Indifinite Integral

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SUMMARY

The integral of the function \(\int \sec^{2}(3x)e^{\tan(3x)}dx\) can be solved using the substitution method. By letting \(u = \tan(3x)\), the integral simplifies to \(\int e^{u} \cdot \sec^{2}(3x)dx\). The differential \(dx\) is expressed as \(dx = \frac{du}{\sec^{2}(3x)} \cdot \frac{1}{3}\), leading to the final solution of \(\frac{1}{3}e^{\tan(3x)} + C\), confirming the correctness of the approach.

PREREQUISITES
  • Understanding of integration techniques, specifically integration by substitution.
  • Familiarity with trigonometric identities, particularly \(\sec^{2}(x) = 1 + \tan^{2}(x)\).
  • Knowledge of exponential functions and their derivatives.
  • Ability to manipulate differentials in calculus.
NEXT STEPS
  • Practice more problems involving integration by substitution with trigonometric functions.
  • Explore the relationship between derivatives and integrals in the context of exponential functions.
  • Study the application of trigonometric identities in calculus.
  • Learn about advanced integration techniques such as integration by parts and partial fractions.
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Students studying calculus, particularly those focusing on integration techniques, as well as educators looking for examples of substitution in integrals.

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Homework Statement



<br /> \int \sec^{2}(3x)e^{\tan(3x)}dx<br />

Homework Equations



The method I am trying to use is integration by substitution(u substitution).



The Attempt at a Solution



I start out by making u = tan(3x)

So i end up having <br /> \int \sec^{2}(3x)e^{u}dx<br />

Stuck after here though. How do I simplify this further?


The final solution as per my professors solution sheet should be <br /> 1/3e^{tan(3x)} + C<br />

However, I cannot seem to figure out why this is the solution.


-Thank you
 
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change sec^{2} 3x into 1 + tan^{2}

Then make substitutions
 
If u=tan(3x), then what's du?
 
Unto said:
change sec^{2} 3x into 1 + tan^{2}

Then make substitutions
This is not a helpful suggestion.
 
Would du be sec^{2}(3x) (3dx)?
 
Yes, it would. Can you use that to finish? Solve that for dx and put it into the original integral.
 
hi, sorry for the late reply.


Solving for dx I got

dx= du/(Sec^2(3x)) x 3

So

Sec^2(3x)e^u du/sec^2(3x) x 3


sec^2(3x) cancel each other out?

so I'm left with

e^u(du)(1/3) = 1/3e^u(du)

so 1/3e^(tan(3x))+C?

Is my approach correct?

-Thank You
 
PolyFX said:
hi, sorry for the late reply.


Solving for dx I got

dx= du/(Sec^2(3x)) x 3

So

Sec^2(3x)e^u du/sec^2(3x) x 3


sec^2(3x) cancel each other out?

so I'm left with

e^u(du)(1/3) = 1/3e^u(du)

so 1/3e^(tan(3x))+C?

Is my approach correct?

-Thank You

Yes, yes, yes.
 

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