Trig Subsitution problem

  • Thread starter teneleven
  • Start date
  • Tags
    Trig
Especially when you go to use LaTeX on a computer algebra system like Maple or Mathematica, it is waiting for dx, otherwise it will give you an answer, but it will be wrong.
  • #1
12
0

Homework Statement



[tex]\int\frac{x^3}{\sqrt{x^2 + 9}}[/tex]

Homework Equations



[tex]x = 3\tan{\theta}[/tex]
[tex]dx=3\sec^2{\theta}[/tex]

The Attempt at a Solution



[tex]27\int\tan^3{\theta}\sec{\theta}[/tex]
[tex]27\int\tan{\theta}(\sec^2{\theta} - 1)\sec{\theta}[/tex]
[tex]27\int(sec^3{\theta} - \sec{\theta})\tan{\theta}[/tex]
[tex]27[\int\sec^3{\theta}\tan{\theta} - \int\sec{\theta}\tan{\theta}[/tex]
[tex]27(\frac{1}{3}\sec^3{\theta} - \sec{\theta})[/tex]
[tex]= 27[\frac{1}{3}(\frac{\sqrt{x^2 + 9}}{3})^3 - \frac{\sqrt{x^2 + 9}}{3}] + C[/tex]

The correct answer is as follows:

[tex]\frac{1}{3}(x^2 - 18)\sqrt{x^2 + 9}[/tex]Any ideas?
 
Last edited:
Physics news on Phys.org
  • #2
Both are correct. You just haven't simplified your answer to the form given in the book.

Btw, Welcome to PF! :)
 
  • #3
Thanks for the welcoming. Love the forums.

I'm trying to simplify it but end up with something different. Where does the -18 come from? I've done the following:

[tex]9(\frac{\sqrt{x^2 + 9}}{3})^3 - \frac{27\sqrt{x^2 + 9}}{3}[/tex]
[tex]9[\frac{(x^2 + 9)\sqrt{x^2 + 9}}{27}] - 9\sqrt{x^2 + 9}[/tex]
[tex]\frac{x^2 + 9\sqrt{x^2 + 9}}{3} - 9\sqrt{x^2 + 9}[/tex]
[tex]\frac{[(x^2 + 9) - 9]\sqrt{x^2 + 9}}{3}[/tex]
[tex]\frac{x^2\sqrt{x^2 + 9}}{3}[/tex]

Thanks.
 
Last edited:
  • #4
[tex]27\left[\frac{\sqrt{x^2+9}}{3}\frac{(x^2+9)}{27} - \frac{\sqrt{x^2+9}}{3}\right]
[/tex]

[tex] = 27\left[\frac{\sqrt{x^2+9}}{3}\left(\frac{(x^2+9)}{27} - 1\right)\right][/tex]
 
Last edited:
  • #5
The first bit of tex you typed, you didn't have a dx stuck on the end. Ill sound pedantic, but you really should not forget them.
 

What is a trig substitution problem?

A trig substitution problem is a mathematical problem that involves using trigonometric identities and functions to solve a more complex integral or equation. It is commonly used in calculus and other advanced math courses.

Why are trig substitution problems important?

Trig substitution problems are important because they allow us to simplify and solve complex integrals and equations that would otherwise be difficult or impossible to solve. They also help us to understand the relationships between different trigonometric functions and how they can be used to manipulate equations.

When should I use a trig substitution?

A trig substitution should be used when you encounter an integral or equation that involves a combination of algebraic and trigonometric functions. In these cases, it can be helpful to substitute the trigonometric functions for their corresponding identities in order to simplify the problem.

What are some common trig substitutions?

Some common trig substitutions include using the identity sin^2(x) + cos^2(x) = 1, which can be used to simplify integrals involving expressions like sqrt(a^2 - x^2) and sqrt(a^2 + x^2). Another common substitution is using the identity sec^2(x) - 1 = tan^2(x), which can be used to simplify integrals involving expressions like sqrt(x^2 - a^2) and sqrt(x^2 + a^2).

How do I know which trig substitution to use?

Choosing the right trig substitution depends on the specific problem you are trying to solve. It is important to carefully examine the integral or equation and identify which trigonometric functions are involved and how they can be simplified using trig identities. Practice and familiarity with different trig substitutions will also help you determine the most appropriate one to use.

Suggested for: Trig Subsitution problem

Replies
7
Views
620
Replies
1
Views
665
Replies
3
Views
886
Replies
3
Views
866
Replies
2
Views
656
Replies
9
Views
1K
Back
Top