# Using integration by parts to prove reduction fomula

1. Apr 17, 2012

### Ianfinity

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.

2. Apr 17, 2012

### Staff: Mentor

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

3. Apr 17, 2012

### Ianfinity

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. Apr 17, 2012

### jimbobian

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.