Stuck in an Integration Problem: Help Needed!

In summary, the conversation is discussing methods for solving the integral \int\sqrt{x^2+9}dx, which has been transformed into the form 3\int\sec^3\theta d\theta. Various suggestions are given, such as using integration by parts and a substitution, and a summary of the steps for integrating \int sec^3(x)dx is also provided.
  • #1
mbrmbrg
496
2
I'm in the middle of solving [tex]\int\sqrt{x^2+9}dx[/tex] and I got it into the form of [tex]3\int\sec^3\theta d\theta[/tex], and I'm pulling a blank. Where does one begin? I could integrate by parts, setting u=sec(theta) and dv=sec^2(theta), but it's not getting me very far. Any suggestions?
 
Physics news on Phys.org
  • #2
bummer. i had that and decided it didn't look appealing...

OK, how's this?

[tex]3\int\sec^3\theta d\theta[/tex]

=[tex]3(\sec\theta\tan\theta-\int\tan^2\theta\sec\theta d\theta)[/tex]

=[tex]3(\sec\theta\tan\theta-\int\frac{sin^2\theta}{cos^3\theta}d\theta)[/tex]

=[tex]3(\sec\theta\tan\theta-\int\frac{1+\cos^2\theta}{cos^3\theta}d\theta)[/tex]

=[tex]3(\sec\theta\tan\theta-\int\frac{1+\cos^2\theta}{cos^3\theta}d\theta)[/tex]

=[tex]3(\sec\theta\tan\theta-\int\sec^3\theta d\theta +\int\sec\theta d\theta)[/tex]

Let I=[tex]\int\sec^3\theta d\theta[/tex]

Then [tex]3I=3(\sec\theta\tan\theta-\int\sec^3\theta d\theta +\int\sec\theta d\theta)[/tex]

The 3's cancel, so [tex]I+I=(\sec\theta\tan\theta+\int\sec\theta d\theta)[/tex]

so [tex]I=(\sec\theta\tan\theta+\int\sec\theta d\theta)[/tex] and I can evaluate it from there (I hope).

Thank you very much!

Where did your post go, Courtigrad? I used it!
 
Last edited:
  • #3
Try the substitution [itex] x=3\sinh t [/itex].

Daniel.
 
  • #4
Here's how I would almost automatically do an integral like that. First convert secant to cosine
[tex]\int sec^3 x dx= \int \frac{1}{cos^3(x)}dx[/itex]
which is an odd power of cosine. "Take out" a cos(x) to use with dx. That is, multiply both numerator and denominator by cos(x)
[tex]\int \frac{cos (x)}{cos^4(x)}dx= \int \frac{cos(x)dx}{(1- sin^2(x))^2}[/tex]
Now let u= sin(x) so du= cos(x)dx
[tex]\int \frac{du}{(1- u^2)^2}= \int \frac{du}{(1-u)^2(1+u)^2}[/tex]
and now use partial fractions.
 
  • #5
Here's how I integrated the [tex]\int sec^3(x)dx[/tex] term.

[tex]\int sec^3(x)dx[/tex]
[tex]=\int sec^2(x)sec(x)[/tex]

[tex]u = sec(x)[/tex]
[tex]du = sec(x)tan(x) dx[/tex]

[tex]dv = sec(x)^2(x)dx[/tex]
[tex]v = tan(x) [/tex][tex]\int sec^3(x)dx[/tex]
[tex]= sec(x)tan(x) - \int tan(x)sec(x)tan(x) dx [/tex]
[tex]= sec(x)tan(x) - \int tan^2(x)sec(x) dx [/tex]
[tex]= sec(x)tan(x) - \int (sec^2(x) - 1)sec(x) dx [/tex]
[tex]= sec(x)tan(x) - \int (sec^3(x) - sec(x) dx [/tex]
[tex]= sec(x)tan(x) - \int (sec^3(x) + \int sec(x) dx [/tex]

Now we have [tex]\int sec^3(x) [/tex] on both sides, so consolidate them.

[tex] 2 \int sec^3(x) dx = sec(x)tan(x) + \int sec(x) dx [/tex]
[tex] 2 \int sec^3(x) dx = sec(x)tan(x) + ln|sec(x) + tan(x)| + K [/tex]
[tex] \int sec^3(x) dx = {{sec(x)tan(x) + ln|sec(x) + tan(x)|}\over{2}} + C [/tex]How did I get
[tex] \int sec(x) dx = ln|sec(x) + tan(x)|[/tex]?

Well, simplify this integral:

[tex] \int sec(x) dx = \int sec(x)*{{sec(x)+tan(x)}\over{sec(x)+tan(x)}}[/tex]
 
Last edited:
  • #6
i would have used the substitution x = 3 tan(t)
 

Related to Stuck in an Integration Problem: Help Needed!

1. What is an integration problem?

An integration problem is a mathematical problem that involves finding the area under a curve or the value of a definite integral. It is often encountered in calculus and is used to solve problems involving rates of change, volumes, and motion.

2. How do you know if you are stuck in an integration problem?

If you are having trouble finding the solution to a problem that requires integrating a function, you may be stuck in an integration problem. This can be identified if you are struggling to manipulate the function or if you are unable to find the right approach to solve the problem.

3. What are some tips for solving integration problems?

First, make sure you understand the basic concepts of integration and have a solid grasp of the fundamental rules and formulas. Then, carefully analyze the problem and try to identify the appropriate technique or method to solve it. Practice and familiarize yourself with common integration techniques, such as substitution, integration by parts, and trigonometric substitutions. Lastly, always double-check your work and make sure to simplify your final answer.

4. What should I do if I am still stuck in an integration problem?

If you are still having trouble, don't hesitate to ask for help. You can consult your teacher or a classmate for guidance. You can also use online resources, such as tutorials or videos, to better understand the concepts and techniques. Additionally, it may be helpful to break down the problem into smaller parts and work through them one at a time.

5. How can I improve my integration problem-solving skills?

Practice is key to improving your integration problem-solving skills. The more problems you solve, the more familiar you will become with different techniques and approaches. You can also challenge yourself by attempting more difficult problems or creating your own problems to solve. It's also helpful to review your work and identify any mistakes or areas for improvement.

Similar threads

  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
914
  • Calculus and Beyond Homework Help
Replies
22
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
34
Views
5K
  • Calculus and Beyond Homework Help
Replies
3
Views
836
  • Calculus and Beyond Homework Help
Replies
3
Views
635
  • Calculus and Beyond Homework Help
Replies
21
Views
1K
Replies
7
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
934
Back
Top