# Thread: Integration by Reduction help! Completly confused.

1. ## Integration by Reduction help! Completly confused.

Alright all im stuck on this integration by reduction question;

$I_n = \int_0^\Pi$ Sin^(2n) $\theta$ Show that

$I_n = \frac{2n-1}{2n} I_(n-1)$ and hence $I_n = \frac{(2n)!}{(2^nn!)^2}$

Sorry my latex isnt so good at the moment.

Help is appricated. Thanks!

2. Originally Posted by simpleas123
Alright all im stuck on this integration by reduction question;

$I_n = \int_0^\Pi Sin^2\cdot n \theta$
I assume you mean

$I_n=\int_{0}^{\pi}\sin^{2n}\theta d\theta$

There are two ways to do this.

First method

Write the integral as for $n \ge 1$

$I_n=\int_{0}^{\pi}\sin\theta \sin^{2n-1}\theta d\theta$

Use integration by parts.

$dv=\sin(\theta)d\theta \implies v=-\cos(\theta)$ and $u=\sin^{2n-1}(\theta) \implies du=(2n-1)\sin^{2n-2}(\theta)\cos(\theta)d\theta$

This gives

$I_n=-\cos(\theta)\sin^{2n-1}(\theta)\bigg|_{0}^{\pi}+(2n-1)\int_{0}^{\pi}\sin^{2n-2}(\theta)\cos^2(\theta)d\theta$

Use the pythagorean Identity from here and you are done

The 2nd is to use complex variables. If you want to see it this way I can show you that as well.

3. Originally Posted by TheEmptySet
I assume you mean

$I_n=\int_{0}^{\pi}\sin^{2n}\theta d\theta$

There are two ways to do this.

First method

Write the integral as for $n \ge 1$

$I_n=\int_{0}^{\pi}\sin\theta \sin^{2n-1}\theta d\theta$

Use integration by parts.

$dv=\sin(\theta)d\theta \implies v=-\cos(\theta)$ and $u=\sin^{2n-1}(\theta) \implies du=(2n-1)\sin^{2n-2}(\theta)\cos(\theta)d\theta$

This gives

$I_n=-\cos(\theta)\sin^{2n-1}(\theta)\bigg|_{0}^{\pi}+(2n-1)\int_{0}^{\pi}\sin^{2n-2}(\theta)\cos^2(\theta)d\theta$

Use the pythagorean Identity from here and you are done

The 2nd is to use complex variables. If you want to see it this way I can show you that as well.

The 1st method is very good, but once i get down to

$I_n = (2n-1)\int_0^\pi$ sin^2n-2 - sin^2n

I dont know were to go from there?

4. Originally Posted by simpleas123
The 1st method is very good, but once i get down to

$I_n = (2n-1)\int_0^\pi$ sin^2n-2 - sin^2n

I dont know were to go from there?
Use the linear property of ingegrals, then ask yourself 2hat is the definition of

$I_{2n} \text{ and } I_{2n-2}$.

Good luck

5. Don't solve this by reduction , use oxidation