Integral Calculus - reduction formula/formulae

[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!

Can anyone help me please!
A kind of walk through lol

Many Thanks

James

 

[Дьявол]

Hi!

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

Here is nice http://archives.math.utk.edu/visual.calculus/4/recursion.2/where explains (using Java or Flash) the derivation of the recursion formulas.

Regards.

 

[rock.freak667]

..

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!

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.

But if we had now

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)

I_n+1= πI_n+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)I_n+2=2^n/2+nI_n

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