Using integration by parts to prove reduction fomula

In summary: In this case, the original integral is divided by (n-1), and then the resulting quotient is multiplied by sec^(n-2).
  • #1
Ianfinity
5
0
Use integration by parts to prove the reduction formula:
int(sec^n)x dx = (tan(x)*sec^(n-2)*x)/(n-1) + [(n-2)/(n-1)]int(sec^(n-2)*x dx

n /= 1 (n does not equal 1)

I used "int" in place of the integral sign.


This was a problem on the corresponding test from the cal A class I am from the past semester. I think there might be something like it on my test tomorrow, but I can't figure it out. Apologies if my formatting of the problem is confusing.

Thanks in advance.
 
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  • #2
Ianfinity said:
Use integration by parts to prove the reduction formula:
int(sec^n)x dx = (tan(x)*sec^(n-2)*x)/(n-1) + [(n-2)/(n-1)]int(sec^(n-2)*x dx

n /= 1 (n does not equal 1)

I used "int" in place of the integral sign.


This was a problem on the corresponding test from the cal A class I am from the past semester. I think there might be something like it on my test tomorrow, but I can't figure it out. Apologies if my formatting of the problem is confusing.

Thanks in advance.

What's your question?

Since you're doing integration by parts, it would help if you show us what you used for u and dv.
 
  • #3
Mark44 said:
What's your question?

Since you're doing integration by parts, it would help if you show us what you used for u and dv.

The question is how do I prove the reduction formula here.

I used u=(sec x)^n and dv=dx, so du=sec(x)tan(x)dx and v=x

Where I'm getting most confused at is the part where it says to add (n-2)/(n-1)int(sec^(n-2))xdx

I don't see why (n-2)/(n-1) is multiplied by the integral there... I know I'm missing something, but what? Do I need to use substitution rule in here at some point?

Thanks for your timely response! I didn't expect to see anything for a few hours.
 
  • #4
Hi Ianfinity,
In your response, you have said that u=(secx)^n , but having made that choice your expression for du/dx is incorrect. I don't think your choice of u is going to yield any useful results, you need to manipulate it a bit before you can apply parts.

As a hint: When you use parts to find a reduction formula, you will usually be aiming to get the new integral to be your original integral to a different power. You can see from the problem statement that this is included in the final answer (sec(x)^(n-2))
You need to think about how the original integral (sec(x)^n) has been split, to allow this term to be formed.
 

What is integration by parts?

Integration by parts is a method used in calculus to find the integral of a product of two functions. It involves using the product rule for differentiation to rewrite the integral in a different form that is easier to solve.

How does integration by parts work?

Integration by parts works by breaking down an integral into two parts and applying the product rule for differentiation to one of the parts. This results in a new integral that is often easier to solve. The process is repeated until the integral can be solved or until a reduction formula is obtained.

What is a reduction formula?

A reduction formula is a formula that allows us to solve integrals of a certain type by reducing them to simpler integrals. It is obtained by repeatedly applying integration by parts to a specific type of integral.

What is the purpose of using integration by parts to prove reduction formulas?

The purpose of using integration by parts to prove reduction formulas is to simplify the process of solving integrals that would otherwise be difficult or impossible to solve. By using integration by parts, we can reduce the integral to a simpler form that can be easily solved using a reduction formula.

What are some common applications of reduction formulas?

Reduction formulas are commonly used in various fields of science, such as physics and engineering, to solve problems involving integrals. They are also used in probability and statistics to calculate probabilities and expected values. Moreover, reduction formulas are useful in solving differential equations and in determining the area under a curve.

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