# Integral Calculus - reduction formula/formulae

1. Aug 25, 2009

### james.farrow

Hi everyone - first a thanks to all who helped me through differential calculus and limits etc - think I'm getting the hang of it....

Anyway as you can probably guess the next topic is integration. I'm kinda stuck on the 'concept' of reduction formula. I've done the usual integration techniques and by parts but this is really difficult!

A kind of walk through lol

Many Thanks

James

2. Aug 25, 2009

### Дьявол

Hi!

Could you possibly explain why do you think its difficult, and what part of reduction formulas don't understand?

Here is nice articlewhere explains (using Java or Flash) the derivation of the recursion formulas.

Regards.

3. Aug 25, 2009

### rock.freak667

The concept is easy, say you had

$$\int_{0} ^{\frac{\pi}{2}} xsinx dx$$

Now you can use integration by parts and get the answer. But as the degree of x goes higher and higher, it becomes tedious to work out as you'd need to apply integration by parts more and more.

$$I_n=\int_0 ^{\frac{\pi}{2}} x^n sinx dx$$

and we expanded that we could somehow get something like that (not correct btw just an example)

In+1= πIn+π/2

so if we wanted to find ∫x3sinx dx or simply I3, we'd have I3=πI2+π/2

and I2=πI1+π/2

we can easily work out I1 (or ∫xsinx dx) as opposed to ∫x3sinx dx

EDIT: Here is a more correct example.

Find

$$\int_0 ^{\frac{\pi}{4}} sec^6 x dx$$

so if we apply a a reduction formula to ∫secnx dx (from π/4 to 0)
we would get a formula of (n+1)In+2=2n/2+nIn

We can find ∫sec x dx from π/4 to 0 or I1. I6 then becomes simple substitution.

Last edited: Aug 25, 2009