Struggling with Integration by Parts? Try a New Approach with Secant Functions!

In summary, integration by parts can be used to solve integrals containing powers of secant, but it is not commonly used for the secant function itself. An alternative method is to use substitution and partial fractions.
  • #1
Ginny Mac
17
0
Integration by parts...

I just started Calc. II and though I struggle a bit, it's fascinating. I have been fooling with a problem lately...one of those standard problems that professors like to assign, and it usually appears in calculus texts:

Have ya'll ever done integration by parts with secx? A friend of mine worked it out for me, but I have had trouble reaching the solution on my own. Just thought I'd throw it out there in case ya'll hadn't run into it! It is a fun kind of headache.

-Gin
 
Physics news on Phys.org
  • #2
Hi Ginny,

I'm scooting this over to our Homework section.

Now, on to your question: Are you talking about integrands that contain powers of [itex]\sec(x)[/itex], or just the [itex]\sec(x)[/itex] itself? The reason I'm asking is that it is normal to integrate odd powers greater than 1 of the secant function by parts, but not so normal to integrate the secant function itself by parts.
 
  • #3
That integral doesn't really require integration by parts... You have to mulitply by a form of one, which can be hard to see.

[tex]\int \sec{x}dx= \int \sec(x)*\frac{\sec(x)+\tan(x)}{\sec(x)+\tan(x)}dx[/tex]

You can do a u-substitution from there.
 
  • #4
Or use the substitution [itex]t = \tan \left( {x/2} \right)[/itex]
 
  • #5
Or [tex]\sec x = \frac{1}{{\cos x}}\left( {\frac{{\cos x}}{{\cos x}}} \right) = \frac{{\cos x}}{{\cos ^2 x}} = \frac{{\cos x}}{{1 - \sin ^2 x}}[/tex] and use substitution + partial fractions. It's one of the longer ways of doing it but it is an alternative. :biggrin:
 

Related to Struggling with Integration by Parts? Try a New Approach with Secant Functions!

1. What is integration by parts and why is it called a headache?

Integration by parts is a technique used in calculus to evaluate integrals that involve products of functions. It is called a headache because it can be a complex and time-consuming process, often leading to confusion and frustration.

2. When should I use integration by parts?

Integration by parts should be used when the integral involves a product of two functions, and the antiderivative of one function is not readily available. It is also useful when the integral involves a function that is difficult to integrate by other methods.

3. What are the steps for integration by parts?

The steps for integration by parts are as follows: 1) Identify the two functions in the integral, 2) Decide which function will be "u" and which will be "dv", 3) Use the formula "u dv = uv - v du" to rewrite the integral, 4) Take the derivative of "u" and the antiderivative of "dv", 5) Substitute the values into the formula and solve for the integral.

4. How do I choose which function to be "u" and which to be "dv"?

There are a few methods for choosing "u" and "dv", including the LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential) and the "ILATE" rule (Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential). It may also depend on the complexity of the functions and trial and error may be necessary.

5. Are there any shortcuts or tricks for integration by parts?

There are some common integrals that can be recognized and solved using integration by parts without going through all the steps. These include "x^n ln(x)", "e^x sin(x)", and "e^x cos(x)". It is also helpful to practice and become familiar with the method to make the process quicker and less daunting.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
1K
Replies
1
Views
3K
  • STEM Academic Advising
Replies
17
Views
2K
  • Calculus and Beyond Homework Help
Replies
9
Views
3K
Replies
11
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
16K
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
  • Calculus and Beyond Homework Help
Replies
7
Views
4K
  • Quantum Physics
Replies
2
Views
1K
Back
Top